DIFFERENT INITIAL AND FINAL HEIGHTS
You throw a ball from your window 8.0 m above the ground.
When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20° below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.
IDENTIFY and SET UP:
As in Examples 3.7 and 3.8, we want to find the horizontal coordinate of a projectile when it is at a given y-value. The difference here is that this value of y is not the same as the initial value. We again choose the x-axis to be horizontal and the y-axis to be upward, and place the origin of coordinates at the point where the ball leaves your hand (Fig. 3.25).
We have v_{0}=10.0 \mathrm{~m} / \mathrm{s} \text { and } \alpha_{0}=-20^{\circ} (the angle is negative because the initial velocity is below the horizontal). Our target variable is the value of x when the ball reaches the ground at y = -8.0 m. We’ll use Eq. (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}) to find the time t when this happens and then use Eq. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t) to find the value of x at this time.
EXECUTE:
To determine t, we rewrite Eq. (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}) in the standard form for a quadratic equation for t:
\frac{1}{2} g t^{2}-\left(v_{0} \sin \alpha_{0}\right) t+y=0The roots of this equation are
t=\frac{v_{0} \sin \alpha_{0} \pm \sqrt{\left(-v_{0} \sin \alpha_{0}\right)^{2}-4\left(\frac{1}{2} g\right) y}}{2\left(\frac{1}{2} g\right)}=\frac{v_{0} \sin \alpha_{0} \pm \sqrt{v_{0}^{2} \sin ^{2} \alpha_{0}-2 g y}}{g}
=\frac{[\begin{array}{l}(10.0 \mathrm{~m} / \mathrm{s}) \sin \left(-20^{\circ}\right) \\\left.{\pm \sqrt{(10.0 \mathrm{~m} / \mathrm{s})^{2} \sin ^{2}\left(-20^{\circ}\right)-2\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(-8.0 \mathrm{~m})}}\right]\end{array}}{9.80 \mathrm{~m} / \mathrm{s}^{2}}
= -1.7 s or 0.98 s
We discard the negative root, since it refers to a time before the ball left your hand. The positive root tells us that the ball reaches the ground at t = 0.98 s. From Eq. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t), the ball’s x-coordinate at that time is
x=\left(v_{0} \cos \alpha_{0}\right) t=(10.0 \mathrm{~m} / \mathrm{s})\left[\cos \left(-20^{\circ}\right)\right](0.98 \mathrm{~s})=9.2 \mathrm{~m}The ball hits the ground a horizontal distance of 9.2 m from your window.
EVALUATE: : The root t = -1.7 s is an example of a “fictional” solution to a quadratic equation. We discussed these in Example 2.8 in Section 2.5; review that discussion.
