how to find max height in projectile motion

Learning Objectives

By the cease of this section, yous will exist able to:

• Use one-dimensional motility in perpendicular directions to clarify projectile movement.
• Calculate the range, time of flight, and maximum pinnacle of a projectile that is launched and impacts a flat, horizontal surface.
• Find the time of flight and impact velocity of a projectile that lands at a different superlative from that of launch.
• Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, field of study only to acceleration as a event of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth'southward atmosphere, fireworks, and the move of any ball in sports. Such objects are called projectiles and their path is chosen a trajectory. The motion of falling objects as discussed in Motion Along a Straight Line is a unproblematic one-dimensional type of projectile motion in which there is no horizontal movement. In this section, nosotros consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The about of import fact to think hither is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Deportation and Velocity Vectors, where nosotros saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis 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 along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the ten-centrality and the vertical axis the y-axis. Information technology is non required that we use this choice of axes; it is simply convenient in the instance of gravitational acceleration. In other cases we may choose a dissimilar set of axes. Figure 4.11 illustrates the annotation for deportation, where we define $\overrightarrow{s}$ to be the total displacement, and $\overrightarrow{x}$ and $\overrightarrow{y}$ are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are southward, ten, and y.

Figure 4.eleven The total displacement s of a soccer ball at a point along its path. The vector $\overrightarrow{s}$ has components $\overrightarrow{x}$ and $\overrightarrow{y}$ along the horizontal and vertical axes. Its magnitude is s and information technology makes an bending Φ with the horizontal.

To describe projectile motion completely, we must include velocity and acceleration, as well every bit deportation. Nosotros must find their components along the x- and y-axes. Let's assume all forces except gravity (such as air resistance and friction, for instance) are negligible. Defining the positive direction to exist upward, the components of acceleration are and so very elementary:

$$a_y=\text{−}\mathrm{chiliad}=\mathrm{-9.eight}\text{m}\text{/}{\text{s}}^2(-32\text{ft}\text{/}{\text{s}}^{\mathrm{two}}).$$

Because gravity is vertical, $a_{\mathrm{ten}}=0.$ If $a_{\mathrm{10}}=0,$ this means the initial velocity in the x direction is equal to the final velocity in the x direction, or $v_x=5_{0x}.$ With these conditions on dispatch and velocity, we can write the kinematic Equation iv.eleven through Equation 4.18 for movement in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration. The kinematic equations for motility in a compatible gravitational field become kinematic equations with $a_y=\text{−}g,a_x=0:$

Horizontal Motion

$$v_{0x}=v_x,x=x_0+v_{\mathrm{10}}t$$

iv.19

Vertical Motion

$$y=y_0+\frac{\mathrm{ane}}{\mathrm{ii}}(5_{0y}+{\mathrm{five}}_y)t$$

four.20

$$y=y_0+v_{0y}t-\frac{\mathrm{i}}{\mathrm{two}}m{t}^2$$

4.22

$$v_y^2=v_{0y}^2-2\mathrm{yard}(y-y_0)$$

4.23

Using this ready of equations, we tin can analyze projectile motion, keeping in heed some important points.

Problem-Solving Strategy

Projectile Motility

1. Resolve the motion into horizontal and vertical components forth the x- and y-axes. The magnitudes of the components of displacement $\overrightarrow{\mathrm{southward}}$ along these axes are 10 and y. The magnitudes of the components of velocity $\overrightarrow{v}$ are $v_x=v\text{cos}\theta \text{and}{v}_y=v\text{sin}\theta ,$ where v is the magnitude of the velocity and θ is its direction relative to the horizontal, every bit shown in Figure 4.12.
2. Treat the motion every bit 2 contained one-dimensional motions: one horizontal and the other vertical. Utilise the kinematic equations for horizontal and vertical motion presented earlier.
3. Solve for the unknowns in the two separate motions: one horizontal and i vertical. Note that the only common variable betwixt the motions is time t. The problem-solving procedures here are the same as those for 1-dimensional kinematics and are illustrated in the post-obit solved examples.
4. Recombine quantities in the horizontal and vertical directions to discover the total displacement $\overrightarrow{s}$ and velocity $\overrightarrow{v}.$ Solve for the magnitude and direction of the displacement and velocity using
   

   $$s=\sqrt{x^{\mathrm{ii}}+y^2},\Phi ={\text{tan}}^{\mathrm{-1}}(y\text{/}\mathrm{ten}),\mathrm{five}=\sqrt{v_{\mathrm{ten}}^2+v_y^{\mathrm{two}}},$$

   
   where Φ is the direction of the displacement $\overrightarrow{s}.$

Effigy 4.12 (a) We analyze two-dimensional projectile move by breaking information technology into 2 independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is unproblematic, because $a_{\mathrm{10}}=0$ and ${\mathrm{five}}_x$ is a constant. (c) The velocity in the vertical direction begins to subtract as the object rises. At its highest betoken, the vertical velocity is cypher. Every bit the object falls toward World again, the vertical velocity increases again in magnitude but points in the reverse direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given indicate on the trajectory.

Example iv.7

A Fireworks Projectile Explodes Loftier and Abroad

During a fireworks display, a trounce is shot into the air with an initial speed of 70.0 yard/s at an angle of $75.0\text{°}$ above the horizontal, as illustrated in Figure iv.13. The fuse is timed to ignite the vanquish but every bit information technology reaches its highest indicate to a higher place the ground. (a) Calculate the height at which the shell explodes. (b) How much fourth dimension passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the crush when it explodes? (d) What is the full deportation from the point of launch to the highest point?

Effigy 4.thirteen The trajectory of a fireworks trounce. The fuse is set to explode the crush at the highest indicate in its trajectory, which is institute to be at a pinnacle of 233 m and 125 m abroad horizontally.

Strategy

The motion tin can be broken into horizontal and vertical motions in which $a_x=0$ and $a_y=\text{−}\mathrm{thousand}.$ We can then define $x_0$ and $y_0$ to be zero and solve for the desired quantities.

Solution

(a) By "acme" nosotros mean the altitude or vertical position y to a higher place the starting bespeak. The highest point in any trajectory, called the apex, is reached when $v_y=0.$ Since nosotros know the initial and final velocities, as well as the initial position, we use the following equation to detect y:

$$v_y^2=v_{0y}^{\mathrm{ii}}-2g(y-y_0).$$

Because $y_0$ and ${\mathrm{five}}_y$ are both zero, the equation simplifies to

$$\text{0}=5_{0y}^{\mathrm{ii}}-2\mathrm{1000}y.$$

Solving for y gives

$$y=\frac{v_{0y}^2}{2g}.$$

At present we must observe $v_{0y},$ the component of the initial velocity in the y direction. It is given past $v_{0y}=v_0\text{sin}{\theta }_0,$ where $v_0$ is the initial velocity of 70.0 thousand/due south and ${\theta }_0=75\text{°}$ is the initial angle. Thus,

$$v_{0y}=5_0\text{sin}\theta =(70.0\text{g}\text{/}\text{southward})\text{sin}75\text{°}=67.6\text{m}\text{/}\text{s}$$

and y is

$$y=\frac{{(67.6\text{m}\text{/}\text{south})}^2}{2(\mathrm{9.fourscore}\text{m}\text{/}{\text{s}}^{\mathrm{ii}})}.$$

Thus, we have

$$y=233\text{m}\text{.}$$

Note that considering up is positive, the initial vertical velocity is positive, equally is the maximum pinnacle, but the acceleration resulting from gravity is negative. Annotation also that the maximum tiptop depends only on the vertical component of the initial velocity, so that whatsoever projectile with a 67.half dozen-m/southward initial vertical component of velocity reaches a maximum pinnacle of 233 thousand (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which exercise accomplish such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would take to exist somewhat larger than that given to accomplish the same pinnacle.

(b) As in many physics problems, at that place is more than i way to solve for the time the projectile reaches its highest betoken. In this example, the easiest method is to use $v_y=v_{0y}-\mathrm{thousand}t.$ Considering $v_y=0$ at the noon, this equation reduces to simply

$$0=v_{0y}-\mathrm{yard}t$$

or

$$t=\frac{v_{0y}}{\mathrm{yard}}=\frac{\mathrm{67.six}\text{m}\text{/}\text{s}}{9.80\text{thousand}\text{/}{\text{due south}}^{\mathrm{two}}}=\mathrm{6.ninety}\text{s}\text{.}$$

This fourth dimension is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass earlier the shell explodes. Another way of finding the time is by using $y\text{=}{y}_0+\frac{\mathrm{i}}{2}(v_{0y}+v_y)t.$ This is left for you as an exercise to consummate.

(c) Because air resistance is negligible, $a_{\mathrm{ten}}=0$ and the horizontal velocity is constant, as discussed before. The horizontal displacement is the horizontal velocity multiplied by time equally given past $\mathrm{10}=x_0+v_{x}t,$ where ${\mathrm{ten}}_0$ is equal to zero. Thus,

$$x=v_{x}t,$$

where $v_{\mathrm{10}}$ is the 10-component of the velocity, which is given by

$$v_x=v_0\text{cos}\theta =(\mathrm{lxx.0}\text{m}\text{/}\text{s})\text{cos}75\text{°}=18.1\text{m}\text{/}\text{s}.$$

Time t for both motions is the same, and so x is

$$x=(18.1\text{m}\text{/}\text{southward})\mathrm{six.90}\text{s}=125\text{m}\text{.}$$

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

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

$$\overrightarrow{s}=125\widehat{i}+233\widehat{j}$$

$$\left|\overrightarrow{s}\right|=\sqrt{{125}^{\mathrm{two}}+{233}^2}=264\text{m}$$

$$\Phi ={\text{tan}}^{\mathrm{-i}}\left(\frac{233}{125}\right)=61.8\text{°}.$$

Notation that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure 4.11, which shows the curvature of the trajectory toward the ground level.

When solving Example 4.7(a), the expression we found for y is valid for whatsoever projectile motion when air resistance is negligible. Phone call the maximum height y = h. And then,

$$h=\frac{{\mathrm{five}}_{0y}^2}{2g}.$$

This equation defines the maximum height of a projectile in a higher place its launch position and it depends just on the vertical component of the initial velocity.

Check Your Agreement 4.three

A stone is thrown horizontally off a cliff $100.0\text{1000}$ high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical movement? (d) What is the stone'south velocity at the point of impact?

Instance 4.viii

Calculating Projectile Motility: Lawn tennis Actor

A tennis role player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle $45\text{°}$ higher up the horizontal (Effigy 4.xiv). On its manner down, the ball is caught by a spectator x m above the point where the ball was hit. (a) Calculate the time it takes the lawn tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball's velocity at impact?

Figure 4.14 The trajectory of a tennis ball hit into the stands.

Strategy

Again, resolving this two-dimensional motion into 2 independent one-dimensional motions allows the states to solve for the desired quantities. The fourth dimension a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t beginning. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This instance asks for the final velocity. Thus, nosotros recombine the vertical and horizontal results to obtain $\overrightarrow{\mathrm{five}}$ at last time t, determined in the first part of the example.

Solution

(a) While the ball is in the air, it rises and then falls to a final position 10.0 m college than its starting altitude. We tin observe the time for this by using Equation 4.22:

$$y\text{=}{y}_0\text{+}{v}_{0y}t-\frac{1}{\mathrm{two}}g{t}^{\mathrm{ii}}.$$

If we take the initial position $y_0$ to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

$$5_{0y}=v_0\text{sin}{\theta }_0=(\mathrm{thirty.0}\text{g}\text{/}\text{s})\text{sin}45\text{°}=21.2\text{m}\text{/}\text{s}.$$

Substituting into Equation iv.22 for y gives us

$$\mathrm{ten.0}\text{1000}=(\mathrm{21.ii}\text{m/s})t-(\mathrm{four.ninety}{\text{m/s}}^{\text{ii}})t^2.$$

Rearranging terms gives a quadratic equation in t:

$$(\mathrm{4.xc}{\text{m/south}}^{\text{two}})t^{\mathrm{two}}-(21.2\text{g/s})t+\mathrm{x.0}\text{g}=0.$$

Utilise of the quadratic formula yields t = 3.79 due south and t = 0.54 s. Since the ball is at a meridian of 10 m at ii times during its trajectory—once on the way up and once on the way down—we accept the longer solution for the time it takes the brawl to accomplish the spectator:

$$t=3.79\text{s}\text{.}$$

The fourth dimension for projectile motion is determined completely past the vertical motion. Thus, whatever projectile that has an initial vertical velocity of 21.2 m/south and lands 10.0 m below its starting distance spends 3.79 south in the air.

(b) We can find the final horizontal and vertical velocities $5_x$ and $v_y$ with the apply of the result from (a). Then, we tin combine them to discover the magnitude of the total velocity vector $\overrightarrow{v}$ and the angle $\theta$ it makes with the horizontal. Since $v_x$ is constant, we tin solve for it at whatever horizontal location. Nosotros choose the starting bespeak because we know both the initial velocity and the initial angle. Therefore,

$${\mathrm{five}}_x=v_0\text{cos}{\theta }_0=(\mathrm{xxx}\text{grand}\text{/}\text{south})\text{cos}45\text{°}=21.2\text{one thousand}\text{/}\text{s}.$$

The terminal vertical velocity is given by Equation four.21:

$$v_y=v_{0y}-gt.$$

Since $v_{0y}$ was plant in role (a) to exist 21.ii one thousand/south, we have

$$5_y=21.2\text{m}\text{/}\text{s}-\mathrm{9.viii}\text{k}\text{/}{\text{south}}^2(\mathrm{iii.79}\text{due south})=\mathrm{-15.9}\text{m}\text{/}\text{southward}.$$

The magnitude of the concluding velocity $\overrightarrow{v}$ is

$$v=\sqrt{{\mathrm{five}}_x^{\mathrm{ii}}+v_y^2}=\sqrt{{(21.2\text{grand}\text{/}\text{s})}^2+{(\text{−}\text{15}\mathrm{.ix}\text{grand}\text{/}\text{s})}^{\mathrm{two}}}=26.5\text{m}\text{/}\text{southward}.$$

The direction ${\theta }_v$ is found using the inverse tangent:

$${\theta }_{\mathrm{five}}={\text{tan}}^{\mathrm{-1}}(\frac{v_y}{5_x})={\text{tan}}^{\mathrm{-one}}\left(\frac{\mathrm{-15.9}}{\mathrm{21.ii}}\right)=\mathrm{36.nine}\text{°}below\; the\; horizon.$$

Significance

(a) Equally mentioned before, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands ten.0 m higher up its starting altitude spends iii.79 s in the air. (b) The negative angle ways the velocity is $\mathrm{36.ix}\text{°}$ below the horizontal at the point of impact. This result is consistent with the fact that the brawl is impacting at a point on the other side of the noon 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 nosotros look since it is impacting 10.0 m in a higher place the launch elevation.

Time of Flight, Trajectory, and Range

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the aforementioned surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flying

We can solve for the fourth dimension of flying of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and deportation in y must be cypher at launch and at impact on an even surface. Thus, we set the deportation in y equal to zero and find

$$y-y_0={\mathrm{five}}_{0y}t-\frac{1}{\mathrm{ii}}\mathrm{grand}{t}^2=(v_0\text{sin}{\theta }_0)t-\frac{1}{\mathrm{ii}}g{t}^2=0.$$

Factoring, we have

$$t\left(v_0\text{sin}{\theta }_0-\frac{gt}{2}\right)=0.$$

Solving for t gives u.s.

$$T_{\text{tof}}=\frac{2(v_0\text{sin}{\theta }_0)}{g}.$$

4.24

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation four.24 does not utilise when the projectile lands at a different elevation than information technology was launched, every bit nosotros saw in Example iv.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the fourth dimension at launch. The fourth dimension 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 World, a projectile launched with the aforementioned velocity as on Earth would exist airborne six times as long.

Trajectory

The trajectory of a projectile tin exist constitute by eliminating the fourth dimension variable t from the kinematic equations for arbitrary t and solving for y(10). Nosotros take $x_0=y_0=0$ and then the projectile is launched from the origin. The kinematic equation for x gives

$$\mathrm{ten}=v_{0x}t\Rightarrow t=\frac{\mathrm{ten}}{v_{0\mathrm{ten}}}=\frac{\mathrm{ten}}{v_0\text{cos}{\theta }_0}.$$

Substituting the expression for t into the equation for the position $y=(v_0\text{sin}{\theta }_0)t-\frac{\mathrm{ane}}{\mathrm{two}}\mathrm{one\; thousand}{t}^2$ gives

$$y=(5_0\text{sin}{\theta }_0)\left(\frac{\mathrm{10}}{v_0\text{cos}{\theta }_0}\right)-\frac{1}{2}g{\left(\frac{x}{v_0\text{cos}{\theta }_0}\right)}^2.$$

Rearranging terms, we take

$$y=(\text{tan}{\theta }_0)x-\left[\frac{\mathrm{grand}}{2{(v_0\text{cos}{\theta }_0)}^2}\right]x^2.$$

4.25

This trajectory equation is of the form $y=a\mathrm{ten}+b{x}^2,$ which is an equation of a parabola with coefficients

$$a=\text{tan}{\theta }_0,b=-\frac{k}{2{(5_0\text{cos}{\theta }_0)}^2}.$$

Range

From the trajectory equation we can also discover the range, or the horizontal distance traveled by the projectile. Factoring Equation 4.25, we have

$$y=\mathrm{ten}\left[\text{tan}{\theta }_0-\frac{g}{2{(v_0\text{cos}{\theta }_0)}^{\mathrm{ii}}}\mathrm{10}\right].$$

The position y is cypher for both the launch bespeak and the touch on point, since nosotros are once again considering merely a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch indicate, and

$$x=\frac{25_0^{\mathrm{two}}\text{sin}{\theta }_0\text{cos}{\theta }_0}{\mathrm{thousand}},$$

corresponding to the impact point. Using the trigonometric identity $2\text{sin}\theta \text{cos}\theta =\text{sin}2\theta$ and setting x = R for range, we notice

$$R=\frac{v_0^{\mathrm{two}}\text{sin}2{\theta }_0}{\mathrm{grand}}.$$

4.26

Note especially that Equation 4.26 is valid only for launch and impact on a horizontal surface. We run into the range is directly proportional to the foursquare of the initial speed $v_0$ and $\text{sin}2{\theta }_0$ , and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would exist six times greater than on Earth for the same initial velocity. Furthermore, we run into from the factor $\text{sin}2{\theta }_0$ that the range is maximum at $45\text{°}.$ These results are shown in Figure iv.15. In (a) we run across that the greater the initial velocity, the greater the range. In (b), nosotros meet that the range is maximum at $45\text{°}.$ This is true only for weather condition neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the aforementioned range is found for ii initial launch angles that sum to $90\text{°}.$ The projectile launched with the smaller angle has a lower apex than the higher angle, just they both have the same range.

Figure 4.fifteen Trajectories of projectiles on level ground. (a) The greater the initial speed $v_0,$ the greater the range for a given initial angle. (b) The event of initial angle ${\theta }_0$ on the range of a projectile with a given initial speed. Note that the range is the same for initial angles of $15\text{°}$ and $75\text{°},$ although the maximum heights of those paths are dissimilar.

Example iv.nine

Comparing Golf Shots

A golfer finds himself in 2 dissimilar situations on different holes. On the second hole he is 120 m from the green and wants to hit the brawl 90 one thousand and let it run onto the green. He angles the shot low to the basis at $30\text{°}$ to the horizontal to let the brawl roll later on impact. On the quaternary hole he is 90 one thousand from the green and wants to let the ball drib with a minimum corporeality of rolling after impact. Here, he angles the shot at $\mathrm{lxx}\text{°}$ to the horizontal to minimize rolling after bear upon. Both shots are hitting and impacted on a level surface.

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

(b) What is the initial speed of the ball at the 4th pigsty?

(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, and then we tin solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

Solution

(a) $R=\frac{v_0^2\text{sin}2{\theta }_0}{g}\Rightarrow 5_0=\sqrt{\frac{Rg}{\text{sin}2{\theta }_0}}=\sqrt{\frac{90.0\text{1000}(\mathrm{nine.8}\text{m}\text{/}{\text{s}}^{\mathrm{two}})}{\text{sin}(2(30\text{°}))}}=31.9\text{one thousand}\text{/}\text{s}$

(b) $R=\frac{v_0^2\text{sin}\mathrm{ii}{\theta }_0}{g}\Rightarrow v_0=\sqrt{\frac{Rg}{\text{sin}2{\theta }_0}}=\sqrt{\frac{90.0\text{chiliad}(9.8\text{m}\text{/}{\text{s}}^{\mathrm{two}})}{\text{sin}(\mathrm{ii}(\mathrm{lxx}\text{°}))}}=37.0\text{m}\text{/}\text{s}$

(c)
$\begin{array}{}\\ \\ \\ y=x\left[\text{tan}{\theta }_0-\frac{\mathrm{thousand}}{\mathrm{ii}{(v_0\text{cos}{\theta }_0)}^2}x\right]\hfill \\ \text{Second hole:}y=x\left[\text{tan}30\text{°}-\frac{9.8\text{m}\text{/}{\text{due south}}^{\mathrm{two}}}{2{[(\mathrm{31.ix}\text{k}\text{/}\text{s)(}\text{cos}30\text{°})]}^2}x\right]=0.58x-0.0064{\mathrm{ten}}^{\mathrm{two}}\hfill \\ \text{Fourth hole:}y=x\left[\text{tan}70\text{°}-\frac{\mathrm{9.viii}\text{m}\text{/}{\text{south}}^2}{2{[(37.0\text{m}\text{/}\text{s)(}\text{cos}70\text{°})]}^2}x\right]=2.75x-0.0306{\mathrm{ten}}^2\hfill \end{array}$

(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 4.16.

Figure iv.16 Two trajectories of a golf game ball with a range of xc 1000. The affect points of both are at the same level as the launch point.

Significance

The initial speed for the shot at $\mathrm{lxx}\text{°}$ is greater than the initial speed of the shot at $30\text{°}.$ Note from Figure 4.16 that if the 2 projectiles were launched at the same speed merely at unlike angles, the projectiles would have the same range every bit long as the angles were less than $90\text{°}.$ The launch angles in this example add together to requite a number greater than $90\text{°}.$ Thus, the shot at $70\text{°}$ has to accept a greater launch speed to accomplish ninety chiliad, otherwise information technology would country at a shorter distance.

Check Your Agreement 4.4

If the two golf game shots in Example 4.9 were launched at the aforementioned speed, which shot would take the greatest range?

When we speak of the range of a projectile on level footing, we assume R is very pocket-size compared with the circumference of Earth. If, however, the range is large, Globe curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, every bit shown in Effigy four.17, which is based on a cartoon in Newton's Principia. If the initial speed is great enough, the projectile goes into orbit. Globe'southward surface drops 5 m every 8000 one thousand. In 1 southward an object falls five one thousand without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/southward (or 18,000 mi/hr) nearly Earth's surface, it will go into orbit around the planet because the surface continuously falls abroad from the object. This is roughly the speed of the Infinite Shuttle in a low Earth orbit when it was operational, or whatever satellite in a low Earth orbit. These and other aspects of orbital motility, such every bit Earth's rotation, are covered in greater depth in Gravitation.

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

Source: https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion

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