HEIGHT AND RANGE OF A PROJECTILE II: MAXIMUM HEIGHT, MAXEMUM RANGE
Find the maximum height h and horizontal range R (see Fig. 3.23) of a projectile launched with speed v_0 at an initial angle \alpha_{0} between 0 and 90°. For a given v_0, what value of \alpha_{0} gives maximum height?
What value gives maximum horizontal range?
IDENTIFY and SET UP:
This is almost the same as parts (b) and (c) of Example 3.7, except that now we want general expressions for h and R. We also want the values of \alpha_{0} that give the maximum values of h and R. In part (b) of Example 3.7 we found that the projectile reaches the high point of its trajectory (so that v_y = 0) at time t_{1}=v_{0 y} / g, and in part (c) we found that the projectile returns to its starting height (so that y = y_0) at time t_{2}=2 v_{0 y} / g=2 t_{1}.
We’ll use Eq. (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}) to find the y-coordinate h at t_1 and Eq. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t) to find the x-coordinate R at time t_2. We’ll express our answers in terms of the launch speed v_0 and launch angle a0 by using Eqs. (3.18) (v_{0 x}=v_{0} \cos \alpha_{0} \quad v_{0 y}=v_{0} \sin \alpha_{0}).
EXECUTE:
From Eqs. (3.18),v_{0 x}=v_{0} \cos \alpha_{0} \text { and } v_{0 y}=v_{0} \sin \alpha_{0}.
Hence we can write the time t_{1} \text { when } v_{y}=0 as
t_{1}=\frac{v_{0 y}}{g}=\frac{v_{0} \sin \alpha_{0}}{g}Equation (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}) gives the height y = h at this time:
h=\left(v_{0} \sin \alpha_{0}\right)\left(\frac{v_{0} \sin \alpha_{0}}{g}\right)-\frac{1}{2} g\left(\frac{v_{0} \sin \alpha_{0}}{g}\right)^{2}=\frac{v_{0}^{2} \sin ^{2} \alpha_{0}}{2 g}For a given launch speed v_0, the maximum value of h occurs for sin \alpha_{0} = 1 and \alpha_{0} = 90°—that is, when the projectile is launched straight up. (If it is launched horizontally, as in Example 3.6, \alpha_{0} = 0 and the maximum height is zero!)
The time t_2 when the projectile hits the ground is
t_{2}=\frac{2 v_{0 y}}{g}=\frac{2 v_{0} \sin \alpha_{0}}{g}The horizontal range R is the value of x at this time. From Eq. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t), this is
R=\left(v_{0} \cos \alpha_{0}\right) t_{2}=\left(v_{0} \cos \alpha_{0}\right) \frac{2 v_{0} \sin \alpha_{0}}{g}=\frac{v_{0}^{2} \sin 2 \alpha_{0}}{g}(We used the trigonometric identity 2 \sin \alpha_{0} \cos \alpha_{0}=\sin 2 \alpha_{0}, found in Appendix B.) The maximum value of \sin 2 \alpha_{0} is 1; this occurs when 2 \alpha_{0}=90^{\circ}, \text { or } \alpha_{0}=45^{\circ}.
This angle gives the maximum range for a given initial speed if air resistance can be ignored.
EVALUATE: Figure 3.24 is based on a composite photograph of three trajectories of a ball projected from a small spring gun at angles of 30°, 45°, and 60°.
The initial speed v_0 is approximately the same in all three cases. The horizontal range is greatest for the 45° angle. The ranges are nearly the same for the 30° and 60° angles: Can you prove that for a given value of v_0 the range is the same for both an initial angle a0 and an initial angle 90° – \alpha_{0}?
(This is not the case in Fig. 3.24 due to air resistance.)
CAUTION :height and range of a projectile We don’t recommend memorizing the above expressions for h and R. They are applicable only in the special circumstances we’ve described. In particular, you can use the expression for the range R only when launch and landing heights are equal. There are many end-of-chapter problems to which these equations do not apply.
