Particle Models in Two Dimensions Projectile Motion Review

4 Motion in Two and Three Dimensions

4.three Projectile Motility

Learning Objectives

By the cease of this section, you will be able to:

• Use i-dimensional motion in perpendicular directions to analyze projectile motion.
• Summate the range, fourth dimension of flight, and maximum height of a projectile that is launched and impacts a apartment, horizontal surface.
• Find the fourth dimension of flight and impact velocity of a projectile that lands at a dissimilar elevation from that of launch.
• Calculate the trajectory of a projectile.

Projectile motion is the motility of an object thrown or projected into the air, subject only to acceleration every bit a result of gravity. The applications of projectile movement in physics and engineering are numerous. Some examples include meteors as they enter Earth'due south atmosphere, fireworks, and the move of any brawl in sports. Such objects are called projectiles and their path is called a trajectory. The motion of falling objects equally discussed in Move Along a Direct Line is a uncomplicated i-dimensional type of projectile motion in which there is no horizontal motion. In this section, nosotros consider 2-dimensional projectile move, and our treatment neglects the furnishings of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus tin exist analyzed separately. Nosotros discussed this fact in Deportation and Velocity Vectors, where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile move is to break it into two motions: one along the horizontal centrality and the other along the vertical. (This choice of axes is the most sensible considering acceleration resulting from gravity is vertical; thus, there is no dispatch forth the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal centrality the x-axis and the vertical axis the y-axis. It is non required that we use this selection of axes; it is simply user-friendly in the case of gravitational acceleration. In other cases we may choose a different set of axes. Effigy illustrates the note for displacement, where we ascertain [latex]\mathbf{\overset{\to }{s}}[/latex] to be the total displacement, and [latex]\mathbf{\overset{\to }{x}}[/latex] and [latex]\mathbf{\overset{\to }{y}}[/latex] are its component vectors forth the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

Figure 4.11 The full deportation s of a soccer brawl at a indicate along its path. The vector [latex]\mathbf{\overset{\to }{s}}[/latex] has components [latex]\mathbf{\overset{\to }{x}}[/latex] and [latex]\mathbf{\overset{\to }{y}}[/latex] along the horizontal and vertical axes. Its magnitude is s and it makes an angle θ with the horizontal.

To draw projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Permit's presume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

[latex]{a}_{y}=\text{−}g=-9.8\,\text{grand}\text{/}{\text{s}}^{ii}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}).[/latex]

Because gravity is vertical, [latex]{a}_{x}=0.[/latex] If [latex]{a}_{x}=0,[/latex] this means the initial velocity in the x direction is equal to the final velocity in the 10 management, or [latex]{v}_{ten}={v}_{0x}.[/latex] With these conditions on dispatch and velocity, we tin write the kinematic (Equation) through (Equation) for motion in a uniform gravitational field, including the rest of the kinematic equations for a abiding acceleration from Motion with Constant Dispatch. The kinematic equations for motility in a uniform gravitational field become kinematic equations with [latex]{a}_{y}=\text{−}g,\enspace{a}_{x}=0:[/latex]

Horizontal Movement

[latex]{v}_{0x}={five}_{ten},\,x={x}_{0}+{v}_{x}t[/latex]

Vertical Move

[latex]y={y}_{0}+\frac{1}{2}({five}_{0y}+{v}_{y})t[/latex]

[latex]{v}_{y}={5}_{0y}-gt[/latex]

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}g{t}^{ii}[/latex]

[latex]{five}_{y}^{2}={v}_{0y}^{two}-2g(y-{y}_{0})[/latex]

Using this set of equations, we tin analyze projectile motion, keeping in heed some of import points.

Problem-Solving Strategy: Projectile Move

1. Resolve the motility into horizontal and vertical components along the ten– and y-axes. The magnitudes of the components of deportation [latex]\mathbf{\overset{\to }{s}}[/latex] along these axes are x and y. The magnitudes of the components of velocity [latex]\mathbf{\overset{\to }{v}}[/latex] are [latex]{five}_{x}=v\text{cos}\,\theta \,\text{and}\,{v}_{y}=5\text{sin}\,\theta ,[/latex] where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Effigy.
2. Treat the move equally ii independent one-dimensional motions: ane horizontal and the other vertical. Utilise the kinematic equations for horizontal and vertical motility presented earlier.
3. Solve for the unknowns in the two separate motions: 1 horizontal and one vertical. Note that the but common variable between the motions is time t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
4. Recombine quantities in the horizontal and vertical directions to find the total displacement [latex]\mathbf{\overset{\to }{south}}[/latex] and velocity [latex]\mathbf{\overset{\to }{v}}.[/latex] Solve for the magnitude and direction of the displacement and velocity using

   [latex]s=\sqrt{{x}^{2}+{y}^{2}},\enspace\theta ={\text{tan}}^{-1}(y\text{/}x),\enspace{v}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}},[/latex]

   where θ is the direction of the deportation [latex]\mathbf{\overset{\to }{south}}.[/latex]

Figure four.12 (a) We analyze ii-dimensional projectile motility by breaking it into two independent i-dimensional motions along the vertical and horizontal axes. (b) The horizontal movement is simple, because [latex]{a}_{x}=0[/latex] and [latex]{v}_{x}[/latex] is a constant. (c) The velocity in the vertical management begins to decrease as the object rises. At its highest point, the vertical velocity is naught. Every bit the object falls toward Earth once more, the vertical velocity increases over again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x and y motions are recombined to give the full velocity at any given bespeak on the trajectory.

Example

A Fireworks Projectile Explodes High and Abroad

During a fireworks brandish, a vanquish is shot into the air with an initial speed of lxx.0 thousand/south at an angle of [latex]75.0^\circ[/latex] above the horizontal, as illustrated in Figure. The fuse is timed to ignite the vanquish merely equally information technology reaches its highest point above the ground. (a) Calculate the height at which the beat out explodes. (b) How much time passes between the launch of the beat out and the explosion? (c) What is the horizontal displacement of the trounce when it explodes? (d) What is the total displacement from the signal of launch to the highest point?

Effigy 4.13 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest betoken in its trajectory, which is establish to be at a summit of 233 m and 125 m away horizontally.

Strategy

The motion can exist broken into horizontal and vertical motions in which [latex]{a}_{x}=0[/latex] and [latex]{a}_{y}=\text{−}chiliad.[/latex] We can then define [latex]{x}_{0}[/latex] and [latex]{y}_{0}[/latex] to be nix and solve for the desired quantities.

Solution

(a) Past "peak" we hateful the altitude or vertical position y in a higher place the starting betoken. The highest point in whatsoever trajectory, called the apex, is reached when [latex]{5}_{y}=0.[/latex] Since we know the initial and final velocities, as well as the initial position, we utilise the following equation to observe y:

[latex]{v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0}).[/latex]

Because [latex]{y}_{0}[/latex] and [latex]{v}_{y}[/latex] are both zero, the equation simplifies to

[latex]\text{0}={five}_{0y}^{2}-2gy.[/latex]

Solving for y gives

[latex]y=\frac{{v}_{0y}^{2}}{2g}.[/latex]

Now we must observe [latex]{v}_{0y},[/latex] the component of the initial velocity in the y direction. It is given by [latex]{5}_{0y}={v}_{0}\text{sin}{\theta }_{0},[/latex] where [latex]{5}_{0}[/latex] is the initial velocity of 70.0 g/southward and [latex]{\theta }_{0}=75^\circ[/latex] is the initial angle. Thus,

[latex]{v}_{0y}={five}_{0}\text{sin}\,\theta =(lxx.0\,\text{g}\text{/}\text{south})\text{sin}\,75^\circ=67.6\,\text{m}\text{/}\text{s}[/latex]

and y is

[latex]y=\frac{{(67.6\,\text{yard}\text{/}\text{southward})}^{2}}{2(9.eighty\,\text{g}\text{/}{\text{southward}}^{2})}.[/latex]

Thus, we take

[latex]y=233\,\text{one thousand}\text{.}[/latex]

Note that because upward is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum tiptop depends merely on the vertical component of the initial velocity, so that whatever projectile with a 67.6-thousand/south initial vertical component of velocity reaches a maximum pinnacle of 233 grand (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do accomplish such heights earlier exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.

(b) As in many physics problems, in that location is more than 1 fashion to solve for the time the projectile reaches its highest point. In this case, the easiest method is to utilize [latex]{v}_{y}={five}_{0y}-gt.[/latex] Considering [latex]{v}_{y}=0[/latex] at the noon, this equation reduces to simply

[latex]0={v}_{0y}-gt[/latex]

or

[latex]t=\frac{{v}_{0y}}{g}=\frac{67.6\,\text{m}\text{/}\text{due south}}{9.lxxx\,\text{m}\text{/}{\text{southward}}^{2}}=6.90\text{due south}\text{.}[/latex]

This time is as well reasonable for big fireworks. If you are able to encounter the launch of fireworks, discover that several seconds pass earlier the shell explodes. Another style of finding the time is by using[latex]y\,\text{=}\,{y}_{0}+\frac{1}{2}({v}_{0y}+{v}_{y})t.[/latex] This is left for you as an do to complete.

(c) Because air resistance is negligible, [latex]{a}_{x}=0[/latex] and the horizontal velocity is constant, as discussed before. The horizontal deportation is the horizontal velocity multiplied past time as given by [latex]x={x}_{0}+{5}_{ten}t,[/latex] where [latex]{x}_{0}[/latex] is equal to zero. Thus,

[latex]x={v}_{10}t,[/latex]

where [latex]{v}_{x}[/latex] is the x-component of the velocity, which is given by

[latex]{v}_{ten}={5}_{0}\text{cos}\,\theta =(70.0\,\text{thou}\text{/}\text{southward})\text{cos}75^\circ=18.i\,\text{m}\text{/}\text{s}.[/latex]

Fourth dimension t for both motions is the aforementioned, so x is

[latex]x=(18.one\,\text{chiliad}\text{/}\text{southward})6.90\,\text{southward}=125\,\text{m}\text{.}[/latex]

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major outcome, and many fragments state directly below.

(d) The horizontal and vertical components of the displacement were only calculated, then all that is needed here is to observe the magnitude and management of the displacement at the highest signal:

[latex]\mathbf{\overset{\to }{s}}=125\mathbf{\chapeau{i}}+233\mathbf{\hat{j}}[/latex]

[latex]|\mathbf{\overset{\to }{s}}|=\sqrt{{125}^{two}+{233}^{2}}=264\,\text{yard}[/latex]

[latex]\theta ={\text{tan}}^{-1}(\frac{233}{125})=61.eight^\circ.[/latex]

Note that the bending for the deportation vector is less than the initial angle of launch. To come across why this is, review Figure, which shows the curvature of the trajectory toward the basis level.

When solving Figure(a), the expression we constitute for y is valid for any projectile motion when air resistance is negligible. Call the maximum peak y = h. Then,

[latex]h=\frac{{v}_{0y}^{2}}{2g}.[/latex]

This equation defines the maximum acme of a projectile above its launch position and it depends merely on the vertical component of the initial velocity.

Bank check Your Agreement

A rock is thrown horizontally off a cliff [latex]100.0\,\text{grand}[/latex] loftier with a velocity of 15.0 grand/s. (a) Ascertain the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical movement? (d) What is the rock's velocity at the point of impact?

Show Solution

(a) Cull the top of the cliff where the rock is thrown from the origin of the coordinate system. Although it is arbitrary, we typically choose fourth dimension t = 0 to correspond to the origin. (b) The equation that describes the horizontal motion is [latex]x={x}_{0}+{five}_{x}t.[/latex] With [latex]{x}_{0}=0,[/latex] this equation becomes [latex]x={v}_{x}t.[/latex] (c) Figure through Figure and Effigy describe the vertical movement, but since [latex]{y}_{0}=0\,\text{and}\,{5}_{0y}=0,[/latex] these equations simplify greatly to become [latex]y=\frac{1}{2}({5}_{0y}+{5}_{y})t=\frac{one}{two}{five}_{y}t,\enspace[/latex] [latex]{5}_{y}=\text{−}gt,\enspace[/latex] [latex]y=-\frac{1}{two}g{t}^{2},\enspace[/latex] and [latex]{v}_{y}^{2}=-2gy.[/latex] (d) We utilise the kinematic equations to find the x and y components of the velocity at the point of impact. Using [latex]{v}_{y}^{2}=-2gy[/latex] and noting the point of impact is −100.0 k, we find the y component of the velocity at impact is [latex]{5}_{y}=44.3\,\text{thou}\text{/}\text{due south}.[/latex] Nosotros are given the x component, [latex]{5}_{10}=fifteen.0\,\text{m}\text{/}\text{south},[/latex] and then we can calculate the total velocity at touch: v = 46.8 m/south and [latex]\theta =71.3^\circ[/latex] beneath the horizontal.

Example

Calculating Projectile Motility: Tennis Thespian

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at thirty thousand/due south and at an angle [latex]45^\circ[/latex] in a higher place the horizontal (Figure). On its way downwardly, the ball is defenseless by a spectator 10 thou above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to accomplish the spectator. (b) What are the magnitude and management of the ball's velocity at impact?

Figure four.14 The trajectory of a tennis brawl hit into the stands.

Strategy

Once again, resolving this two-dimensional move into 2 independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed past its vertical motility lonely. Thus, we solve for t commencement. While the ball is rise and falling vertically, the horizontal motility continues at a abiding velocity. This case asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain [latex]\mathbf{\overset{\to }{v}}[/latex] at final time t, determined in the starting time part of the instance.

Solution

(a) While the ball is in the air, information technology rises and and so falls to a final position 10.0 yard higher than its starting distance. We tin notice the fourth dimension for this past using Effigy:

[latex]y\,\text{=}\,{y}_{0}\,\text{+}\,{5}_{0y}t-\frac{1}{2}g{t}^{2}.[/latex]

If we take the initial position [latex]{y}_{0}[/latex] to exist nothing, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

[latex]{v}_{0y}={5}_{0}\text{sin}\,{\theta }_{0}=(thirty.0\,\text{m}\text{/}\text{due south})\text{sin}\,45^\circ=21.2\,\text{m}\text{/}\text{s}.[/latex]

Substituting into Effigy for y gives us

[latex]10.0\,\text{thou}=(21.2\,\text{m/s})t-(4.90\,{\text{m/s}}^{\text{2}}){t}^{2}.[/latex]

Rearranging terms gives a quadratic equation in t:

[latex](4.xc\,{\text{1000/southward}}^{\text{2}}){t}^{two}-(21.ii\,\text{m/s})t+ten.0\,\text{m}=0.[/latex]

Use of the quadratic formula yields t = iii.79 southward and t = 0.54 s. Since the ball is at a superlative of 10 m at ii times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time information technology takes the ball to attain the spectator:

[latex]t=three.79\,\text{s}\text{.}[/latex]

The time for projectile move is determined completely by the vertical motility. Thus, any projectile that has an initial vertical velocity of 21.2 one thousand/s and lands 10.0 1000 below its starting altitude spends 3.79 s in the air.

(b) We can find the final horizontal and vertical velocities [latex]{v}_{ten}[/latex] and [latex]{v}_{y}[/latex] with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector [latex]\mathbf{\overset{\to }{v}}[/latex] and the bending [latex]\theta[/latex] information technology makes with the horizontal. Since [latex]{5}_{ten}[/latex] is abiding, we can solve for information technology at any horizontal location. We choose the starting signal because we know both the initial velocity and the initial angle. Therefore,

[latex]{v}_{x}={five}_{0}\text{cos}{\theta }_{0}=(30\,\text{m}\text{/}\text{s})\text{cos}\,45^\circ=21.2\,\text{m}\text{/}\text{s}.[/latex]

The final vertical velocity is given by Figure:

[latex]{five}_{y}={v}_{0y}-gt.[/latex]

Since [latex]{v}_{0y}[/latex] was found in part (a) to be 21.2 k/south, we take

[latex]{v}_{y}=21.2\,\text{m}\text{/}\text{s}-9.eight\,\text{1000}\text{/}{\text{s}}^{2}(3.79\,\text{s})=-15.9\,\text{m}\text{/}\text{s}.[/latex]

The magnitude of the concluding velocity [latex]\mathbf{\overset{\to }{v}}[/latex] is

[latex]five=\sqrt{{v}_{x}^{ii}+{v}_{y}^{2}}=\sqrt{{(21.2\,\text{m}\text{/}\text{southward})}^{ii}+{(\text{−}\,\text{xv}.9\,\text{g}\text{/}\text{s})}^{2}}=26.5\,\text{m}\text{/}\text{s}.[/latex]

The direction [latex]{\theta }_{v}[/latex] is found using the inverse tangent:

[latex]{\theta }_{v}={\text{tan}}^{-ane}(\frac{{v}_{y}}{{five}_{x}})={\text{tan}}^{-one}(\frac{21.two}{-15.nine})=-53.ane^\circ.[/latex]

Significance

(a) As mentioned earlier, the time for projectile motion is adamant completely past the vertical motility. Thus, any projectile that has an initial vertical velocity of 21.2 m/south and lands 10.0 m beneath its starting distance spends 3.79 s in the air. (b) The negative angle means the velocity is [latex]53.1^\circ[/latex] below the horizontal at the betoken of impact. This outcome is consistent with the fact that the ball is impacting at a bespeak on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we await since it is impacting 10.0 m above the launch elevation.

Time of Flying, Trajectory, and Range

Of interest are the fourth dimension of flying, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this instance, kinematic equations requite useful expressions for these quantities, which are derived in the following sections.

Time of flying

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We annotation the position and deportation in y must exist null at launch and at bear on on an fifty-fifty surface. Thus, we set the deportation in y equal to zero and find

[latex]y-{y}_{0}={v}_{0y}t-\frac{i}{ii}g{t}^{two}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{ii}g{t}^{2}=0.[/latex]

Factoring, nosotros have

[latex]t({v}_{0}\text{sin}{\theta }_{0}-\frac{gt}{ii})=0.[/latex]

Solving for t gives us

[latex]{T}_{\text{tof}}=\frac{ii({5}_{0}\text{sin}{\theta }_{0})}{g}.[/latex]

This is the time of flying for a projectile both launched and impacting on a flat horizontal surface. Figure does not apply when the projectile lands at a different acme than it was launched, every bit we saw in Figure of the tennis role player striking the brawl into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is 1-sixth that of Earth, a projectile launched with the same velocity equally on World would be airborne six times as long.

Trajectory

The trajectory of a projectile can be constitute by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(10). We have [latex]{x}_{0}={y}_{0}=0[/latex] and so the projectile is launched from the origin. The kinematic equation for ten gives

[latex]10={five}_{0x}t\Rightarrow t=\frac{x}{{v}_{0x}}=\frac{x}{{five}_{0}\text{cos}{\theta }_{0}}.[/latex]

Substituting the expression for t into the equation for the position [latex]y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{ane}{two}g{t}^{2}[/latex] gives

[latex]y=({v}_{0}\text{sin}\,{\theta }_{0})(\frac{x}{{5}_{0}\text{cos}\,{\theta }_{0}})-\frac{one}{2}g{(\frac{x}{{v}_{0}\text{cos}\,{\theta }_{0}})}^{2}.[/latex]

Rearranging terms, nosotros have

[latex]y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({5}_{0}\text{cos}\,{\theta }_{0})}^{two}}]{x}^{2}.[/latex]

This trajectory equation is of the form [latex]y=ax+b{x}^{2},[/latex] which is an equation of a parabola with coefficients

[latex]a=\text{tan}\,{\theta }_{0},\enspace b=-\frac{one thousand}{2{({5}_{0}\text{cos}\,{\theta }_{0})}^{2}}.[/latex]

Range

From the trajectory equation nosotros can also discover the range, or the horizontal distance traveled by the projectile. Factoring Effigy, nosotros have

[latex]y=x[\text{tan}\,{\theta }_{0}-\frac{g}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x].[/latex]

The position y is nil for both the launch point and the impact point, since we are again considering just a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch betoken, and

[latex]ten=\frac{2{v}_{0}^{ii}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{thousand},[/latex]

corresponding to the impact point. Using the trigonometric identity [latex]2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}ii\theta[/latex] and setting x = R for range, we discover

[latex]R=\frac{{v}_{0}^{2}\text{sin}2{\theta }_{0}}{g}.[/latex]

Note particularly that Figure is valid only for launch and impact on a horizontal surface. We meet the range is directly proportional to the square of the initial speed [latex]{v}_{0}[/latex] and [latex]\text{sin}2{\theta }_{0}[/latex], and it is inversely proportional to the dispatch of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, nosotros encounter from the factor [latex]\text{sin}2{\theta }_{0}[/latex] that the range is maximum at [latex]45^\circ.[/latex] These results are shown in Effigy. In (a) nosotros see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at [latex]45^\circ.[/latex] This is true but for atmospheric condition neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. Information technology is interesting that the same range is found for two initial launch angles that sum to [latex]90^\circ.[/latex] The projectile launched with the smaller angle has a lower apex than the college bending, but they both have the same range.

Figure iv.fifteen Trajectories of projectiles on level footing. (a) The greater the initial speed [latex]{v}_{0},[/latex] the greater the range for a given initial bending. (b) The outcome of initial angle [latex]{\theta }_{0}[/latex] on the range of a projectile with a given initial speed. Note that the range is the same for initial angles of [latex]fifteen^\circ[/latex] and [latex]75^\circ,[/latex] although the maximum heights of those paths are dissimilar.

Example

Comparison Golf game Shots

A golfer finds himself in two unlike situations on dissimilar holes. On the second pigsty he is 120 1000 from the green and wants to striking the ball 90 thou and let it run onto the dark-green. He angles the shot depression to the ground at [latex]thirty^\circ[/latex] to the horizontal to let the ball scroll after touch. On the 4th hole he is 90 g from the green and wants to let the brawl drib with a minimum corporeality of rolling after impact. Here, he angles the shot at [latex]70^\circ[/latex] to the horizontal to minimize rolling later on impact. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the ball at the second hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

Strategy

We encounter that the range equation has the initial speed and angle, then we tin solve for the initial speed for both (a) and (b). When we accept the initial speed, nosotros tin apply this value to write the trajectory equation.

Solution

(a) [latex]R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{k}\Rightarrow {v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,two{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(nine.viii\,\text{grand}\text{/}{\text{south}}^{two})}{\text{sin}(ii(seventy^\circ))}}=37.0\,\text{m}\text{/}\text{s}[/latex]

(b) [latex]R=\frac{{5}_{0}^{ii}\text{sin}\,2{\theta }_{0}}{g}\Rightarrow {5}_{0}=\sqrt{\frac{Rg}{\text{sin}\,two{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{yard}(ix.eight\,\text{k}\text{/}{\text{s}}^{ii})}{\text{sin}(ii(30^\circ))}}=31.9\,\text{m}\text{/}\text{due south}[/latex]

(c)

[latex]\begin{array}{cc} y=10[\text{tan}\,{\theta }_{0}-\frac{g}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{2nd pigsty:}\,y=10[\text{tan}\,70^\circ-\frac{9.8\,\text{m}\text{/}{\text{due south}}^{2}}{2{[(37.0\,\text{g}\text{/}\text{s)(}\text{cos}\,70^\circ)]}^{2}}x]=ii.75x-0.0306{x}^{2}\hfill \\ \text{Quaternary pigsty:}\,y=x[\text{tan}\,30^\circ-\frac{ix.8\,\text{m}\text{/}{\text{southward}}^{two}}{ii{[(31.9\,\text{grand}\text{/}\text{due south)(}\text{cos}30^\circ)]}^{2}}x]=0.58x-0.0064{x}^{two}\hfill \finish{array}[/latex]

(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure.

Figure 4.16 2 trajectories of a golf ball with a range of 90 chiliad. The impact points of both are at the aforementioned level as the launch betoken.

Significance

The initial speed for the shot at [latex]lxx^\circ[/latex] is greater than the initial speed of the shot at [latex]thirty^\circ.[/latex] Note from Figure that ii projectiles launched at the same speed just at unlike angles accept the aforementioned range if the launch angles add together to [latex]90^\circ.[/latex] The launch angles in this example add to requite a number greater than [latex]90^\circ.[/latex] Thus, the shot at [latex]seventy^\circ[/latex] has to have a greater launch speed to accomplish 90 m, otherwise information technology would land at a shorter distance.

Check Your Understanding

If the two golf game shots in Figure were launched at the aforementioned speed, which shot would have the greatest range?

Evidence Solution

The golf game shot at [latex]30^\circ.[/latex]

When we speak of the range of a projectile on level basis, we presume R is very minor compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction forth the path. The range is larger than predicted by the range equation given earlier because the projectile has further to fall than it would on level ground, as shown in Effigy, which is based on a drawing in Newton'due south Principia . If the initial speed is smashing enough, the projectile goes into orbit. Earth's surface drops v m every 8000 chiliad. In 1 s an object falls five one thousand without air resistance. Thus, if an object is given a horizontal velocity of [latex]8000\,\text{grand}\text{/}\text{s}[/latex] (or [latex]eighteen,000\text{mi}\text{/}\text{60 minutes})[/latex] near Globe'south surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when information technology was operational, or any satellite in a low Earth orbit. These and other aspects of orbital move, such as Earth'south rotation, are covered in greater depth in Gravitation.

Figure iv.17 Projectile to satellite. In each case shown here, a projectile is launched from a very loftier tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level basis because Earth curves abroad beneath its path. With a speed of 8000 m/s, orbit is accomplished.

Summary

• Projectile motion is the motion of an object bailiwick but to the acceleration of gravity, where the acceleration is constant, as near the surface of Earth.
• To solve projectile move issues, we analyze the motion of the projectile in the horizontal and vertical directions using the one-dimensional kinematic equations for x and y.
• The time of flight of a projectile launched with initial vertical velocity [latex]{v}_{0y}[/latex] on an even surface is given by

  [latex]{T}_{tof}=\frac{2({five}_{0}\text{sin}\,\theta )}{g}.[/latex]

  This equation is valid but when the projectile lands at the same elevation from which it was launched.

• The maximum horizontal distance traveled by a projectile is called the range. Again, the equation for range is valid only when the projectile lands at the same elevation from which information technology was launched.

Conceptual Questions

Respond the following questions for projectile motion on level ground bold negligible air resistance, with the initial angle being neither [latex]0^\circ[/latex] nor [latex]90^\circ:[/latex] (a) Is the velocity always zero? (b) When is the velocity a minimum? A maximum? (c) Tin can the velocity e'er be the same as the initial velocity at a time other than at t = 0? (d) Tin the speed always be the same as the initial speed at a fourth dimension other than at t = 0?

Evidence Solution

a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yep, where it lands

Answer the following questions for projectile motion on level ground assuming negligible air resistance, with the initial bending being neither [latex]0^\circ[/latex] nor [latex]xc^\circ:[/latex] (a) Is the acceleration ever naught? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the dispatch always reverse in management to a component of velocity?

A dime is placed at the edge of a tabular array and so it hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime head on. Which coin hits the ground first?

Show Solution

They both hitting the ground at the same time.

Problems

A bullet is shot horizontally from shoulder height (i.five g) with and initial speed 200 m/due south. (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

Show Solution

a. [latex]t=0.55\,\text{s}[/latex], b. [latex]x=110\,\text{m}[/latex]

A marble rolls off a tabletop 1.0 m high and hits the floor at a signal 3.0 m away from the tabular array'southward edge in the horizontal management. (a) How long is the marble in the air? (b) What is the speed of the marble when it leaves the table's edge? (c) What is its speed when information technology hits the floor?

A dart is thrown horizontally at a speed of 10 one thousand/s at the bull's-eye of a dartboard ii.four m away, as in the following effigy. (a) How far below the intended target does the dart hit? (b) What does your answer tell you lot about how proficient dart players throw their darts?

Show Solution

a. [latex]t=0.24\text{s,}\enspace d=0.28\,\text{thou}[/latex], b. They aim high.

An airplane flying horizontally with a speed of 500 km/h at a height of 800 m drops a crate of supplies (see the following effigy). If the parachute fails to open, how far in front of the release betoken does the crate hit the ground?

Suppose the airplane in the preceding problem fires a projectile horizontally in its direction of motion at a speed of 300 m/southward relative to the aeroplane. (a) How far in front of the release point does the projectile hit the ground? (b) What is its speed when information technology hits the footing?

Evidence Solution

a., [latex]t=12.eight\,\text{s,}\enspace x=5619\,\text{grand}[/latex] b. [latex]{five}_{y}=125.0\,\text{m}\text{/}\text{south,}\enspace{five}_{x}=439.0\,\text{grand}\text{/}\text{s,}\enspace |\mathbf{\overset{\to }{v}}|=456.0\,\text{m}\text{/}\text{due south}[/latex]

A fastball pitcher tin throw a baseball at a speed of 40 thou/south (90 mi/h). (a) Assuming the bullpen can release the ball 16.7 k from habitation plate and so the ball is moving horizontally, how long does information technology take the brawl to achieve home plate? (b) How far does the ball drop betwixt the pitcher'south hand and home plate?

A projectile is launched at an angle of [latex]30^\circ[/latex] and lands 20 due south later at the same height equally information technology was launched. (a) What is the initial speed of the projectile? (b) What is the maximum distance? (c) What is the range? (d) Calculate the displacement from the point of launch to the position on its trajectory at xv southward.

Show Solution

a. [latex]{five}_{y}={v}_{0y}-gt,\enspace t=10\text{southward,}\enspace{v}_{y}=0,\enspace{5}_{0y}=98.0\,\text{yard}\text{/}\text{southward},\enspace{five}_{0}=196.0\,\text{m}\text{/}\text{s}[/latex], b. [latex]h=490.0\,\text{chiliad},[/latex]

c. [latex]{v}_{0x}=169.7\,\text{yard}\text{/}\text{southward,}\enspace x=3394.0\,\text{one thousand,}[/latex]

d. [latex]\begin{assortment}{cc} x=2545.5\,\text{one thousand}\hfill \\ y=465.5\,\text{m}\hfill \\ \mathbf{\overset{\to }{s}}=2545.5\,\text{m}\mathbf{\chapeau{i}}+465.five\,\text{m}\mathbf{\hat{j}}\hfill \end{array}[/latex]

A basketball game player shoots toward a handbasket 6.one k away and 3.0 m to a higher place the floor. If the ball is released 1.viii one thousand to a higher place the floor at an angle of [latex]60^\circ[/latex] to a higher place the horizontal, what must the initial speed exist if it were to go through the handbasket?

At a detail instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 yard/southward. At this exact instant, a daughter throws a ball horizontally, relative to herself, with an initial speed of 20 m/south. When she lands, where will she find the ball? Ignore air resistance.

Show Solution

[latex]-100\,\text{m}=(-2.0\,\text{m}\text{/}\text{s})t-(4.9\,\text{m}\text{/}{\text{s}}^{2}){t}^{two},[/latex] [latex]t=4.3\,\text{due south,}[/latex] [latex]ten=86.0\,\text{m}[/latex]

A man on a motorbike traveling at a compatible speed of 10 m/south throws an empty can straight upwards relative to himself with an initial speed of three.0 yard/s. Find the equation of the trajectory as seen by a constabulary officeholder on the side of the road. Presume the initial position of the can is the bespeak where it is thrown. Ignore air resistance.

An athlete tin can jump a altitude of eight.0 m in the broad spring. What is the maximum distance the athlete tin spring on the Moon, where the gravitational dispatch is 1-sixth that of Earth?

Show Solution

[latex]{R}_{Moon}=48\,\text{m}[/latex]

The maximum horizontal distance a boy can throw a ball is 50 thousand. Assume he tin throw with the same initial speed at all angles. How high does he throw the ball when he throws it directly upwardly?

A rock is thrown off a cliff at an bending of [latex]53^\circ[/latex] with respect to the horizontal. The cliff is 100 grand high. The initial speed of the rock is 30 thou/s. (a) How high above the edge of the cliff does the stone rise? (b) How far has information technology moved horizontally when it is at maximum distance? (c) How long after the release does it striking the ground? (d) What is the range of the rock? (east) What are the horizontal and vertical positions of the stone relative to the edge of the cliff at t = 2.0 s, t = 4.0 s, and t = 6.0 s?

Bear witness Solution

a.[latex]{v}_{0y}=24\,\text{1000}\text{/}\text{s}[/latex] [latex]{v}_{y}^{two}={v}_{0y}^{2}-2gy\Rightarrow h=23.4\,\text{m}[/latex],

b. [latex]t=3\,\text{s}\enspace{five}_{0x}=18\,\text{one thousand/s}\enspace ten=54\,\text{thou}[/latex],

c. [latex]y=-100\,\text{m}\enspace{y}_{0}=0[/latex] [latex]y-{y}_{0}={v}_{0y}t-\frac{ane}{two}g{t}^{2}\enspace -100=24t-4.9{t}^{2}[/latex] [latex]\Rightarrow t=seven.58\,\text{s}[/latex],

d. [latex]x=136.44\,\text{m}[/latex],

e. [latex]t=two.0\,\text{s}\enspace y=28.four\,\text{chiliad}\enspace x=36\,\text{yard}[/latex]

[latex]t=4.0\,\text{south}\enspace y=17.half dozen\,\text{m}\enspace 10=22.four\,\text{1000}[/latex]

[latex]t=six.0\,\text{southward}\enspace y=-32.4\,\text{m}\enspace 10=108\,\text{chiliad}[/latex]

Trying to escape his pursuers, a secret agent skis off a slope inclined at [latex]30^\circ[/latex] beneath the horizontal at lx km/h. To survive and state on the snow 100 m below, he must clear a gorge 60 m wide. Does he make it? Ignore air resistance.

A golfer on a fairway is lxx m away from the light-green, which sits beneath the level of the fairway by 20 m. If the golfer hits the ball at an angle of [latex]40^\circ[/latex] with an initial speed of 20 m/southward, how shut to the green does she come?

Show Solution

[latex]{5}_{0y}=12.nine\,\text{chiliad}\text{/}\text{s}\,y-{y}_{0}={five}_{0y}t-\frac{1}{two}g{t}^{2}\enspace -20.0=12.9t-4.9{t}^{2}[/latex]

[latex]t=3.7\,\text{s}\enspace{v}_{0x}=15.3\,\text{m}\text{/}\text{s}\Rightarrow ten=56.7\,\text{m}[/latex]

So the golfer's shot lands xiii.3 chiliad curt of the green.

A projectile is shot at a loma, the base of which is 300 m away. The projectile is shot at [latex]60^\circ[/latex] above the horizontal with an initial speed of 75 m/s. The hill can exist approximated by a plane sloped at [latex]20^\circ[/latex] to the horizontal. Relative to the coordinate system shown in the following figure, the equation of this direct line is [latex]y=(\text{tan}20^\circ)ten-109.[/latex] Where on the loma does the projectile land?

An astronaut on Mars kicks a soccer ball at an angle of [latex]45^\circ[/latex] with an initial velocity of fifteen m/due south. If the acceleration of gravity on Mars is 3.seven chiliad/due south, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the aforementioned kick on the Moon, where gravity is 1-sixth that of Earth?

Prove Solution

a. [latex]R=60.viii\,\text{thou}[/latex],

b. [latex]R=137.8\,\text{1000}[/latex]

Mike Powell holds the tape for the long jump of viii.95 thou, established in 1991. If he left the ground at an angle of [latex]15^\circ,[/latex] what was his initial speed?

MIT's robot cheetah can jump over obstacles 46 cm loftier and has speed of 12.0 km/h. (a) If the robot launches itself at an angle of [latex]sixty^\circ[/latex] at this speed, what is its maximum superlative? (b) What would the launch angle have to exist to attain a height of 46 cm?

Show Solution

a. [latex]{v}_{y}^{2}={5}_{0y}^{ii}-2gy\Rightarrow y=2.nine\,\text{thousand}\text{/}\text{s}[/latex]

[latex]y=three.3\,\text{m}\text{/}\text{s}[/latex]

[latex]y=\frac{{v}_{0y}^{ii}}{2g}=\frac{{({five}_{0}\text{sin}\,\theta )}^{two}}{2g}\Rightarrow \text{sin}\,\theta =0.91\Rightarrow \theta =65.5^\circ[/latex]

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an angle of [latex]forty^\circ[/latex] with respect to the horizontal and landed 900 yard beneath the crater, (a) what would be their initial velocity and (b) what is their time of flying?

Drew Brees of the New Orleans Saints can throw a football 23.0 yard/s (50 mph). If he angles the throw at [latex]x^\circ[/latex] from the horizontal, what distance does it become if information technology is to exist caught at the same elevation as it was thrown?

Show Solution

[latex]R=18.5\,\text{m}[/latex]

The Lunar Roving Vehicle used in NASA's belatedly Apollo missions reached an unofficial lunar land speed of 5.0 thousand/s by astronaut Eugene Cernan. If the rover was moving at this speed on a apartment lunar surface and hit a modest crash-land that projected it off the surface at an angle of [latex]20^\circ,[/latex] how long would it be "airborne" on the Moon?

A soccer goal is 2.44 thou high. A player kicks the ball at a distance 10 grand from the goal at an angle of [latex]25^\circ.[/latex] What is the initial speed of the soccer ball?

Show Solution

[latex]y=(\text{tan}\,{\theta }_{0})ten-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}\Rightarrow {5}_{0}=16.iv\,\text{chiliad}\text{/}\text{s}[/latex]

Olympus Mons on Mars is the largest volcano in the solar organisation, at a top of 25 km and with a radius of 312 km. If you are standing on the superlative, with what initial velocity would yous take to fire a projectile from a cannon horizontally to articulate the volcano and land on the surface of Mars? Notation that Mars has an dispatch of gravity of [latex]3.7\,\text{one thousand}\text{/}{\text{s}}^{two}.[/latex]

In 1999, Robbie Knievel was the first to bound the M Coulee on a motorbike. At a narrow part of the canyon (69.0 m wide) and traveling 35.eight m/s off the takeoff ramp, he reached the other side. What was his launch angle?

Evidence Solution

[latex]R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{thousand}\Rightarrow {\theta }_{0}=15.0^\circ[/latex]

You throw a baseball game at an initial speed of 15.0 m/southward at an angle of [latex]30^\circ[/latex] with respect to the horizontal. What would the brawl's initial speed accept to be at [latex]30^\circ[/latex] on a planet that has twice the dispatch of gravity equally Globe to reach the same range? Consider launch and impact on a horizontal surface.

Aaron Rogers throws a football at twenty.0 g/due south to his wide receiver, who runs direct down the field at nine.4 yard/s for 20.0 k. If Aaron throws the football when the broad receiver has reached 10.0 one thousand, what bending does Aaron have to launch the ball so the receiver catches it at the 20.0 m mark?

Prove Solution

It takes the wide receiver 1.ane s to comprehend the last 10 yard of his run.

[latex]{T}_{\text{tof}}=\frac{two({v}_{0}\text{sin}\,\theta )}{g}\Rightarrow \text{sin}\,\theta =0.27\Rightarrow \theta =15.vi^\circ[/latex]

Glossary

projectile motion

motion of an object subject merely to the acceleration of gravity

range

maximum horizontal distance a projectile travels

time of flight

elapsed time a projectile is in the air

trajectory

path of a projectile through the air

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Source: https://pressbooks.online.ucf.edu/phy2048tjb/chapter/4-3-projectile-motion/

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