
(1) Find the time t_1 from A to C.
(2) Find the height h_1.
(3) After colliding the wall, find the time t_2 when reaching the highest point in the projectile D.
(4) Find the height h_2.
(5) Find the time t_3 from the highest position of projectile D to A.
(6) Find the coefficient of restitution e.
Answer:
(1) The initial velocity in the horizontal direction is expressed as v_0\cos\theta, which is constant through the motion. Therefore, the range R is expressed by v_0\cos\theta \times t_1, so
t_1 = \frac{R}{v_0\cos\theta}
(2) Consider the vertical direction. The initial velocity is v_0\sin\theta and the time is given in the above.
\begin{eqnarray*} h_1 &=& v_0\sin\theta t_1 - \frac{1}{2}gt_1^2 \\ &=& v_0\sin\theta \frac{R}{v_0\cos\theta} - \frac{1}{2}g\left(\frac{R}{v_0\cos\theta}\right)^2\\ &=& R\tan\theta - \frac{1}{2}g\left(\frac{R}{v_0\cos\theta}\right)^2 \end{eqnarray*}
(3) Let's consider the motion from A to the highest point of D. The time is defined as t'. Use the kinematic equation.
v_y = v_0\sin\theta -gt'
Since v_y=0 at the highest point, we obtain t' as follows.
gt' = v_0\sin\theta \quad \rightarrow \quad t' = \frac{v_0\sin\theta}{g}
However, t_2 is from C, so subtract t_1 from t'.
t_2 = t' - t_1 = \frac{v_0\sin\theta}{g} - \frac{R}{v_0\cos\theta}
(4) The height h_2 is given by the kinematic equation.
h_2 = v_0\sin\theta t' - \frac{1}{2}gt'^2 \\ = \frac{(v_0\sin\theta)^2}{g}-\frac{1}{2}\left(\frac{v_0\sin\theta}{g}\right) \\ = \frac{v_0\sin\theta (2v_0\sin\theta-1)}{2g}
(5) The motion is symmetric, so
t_3 = t'
(6) Define the coefficient of restitution by the distances before and after collision; namely, we have
e = \frac{\mbox{distance before collision}}{\mbox{distance after collision}}
The distance before collision is R. The distance after collision is that the total time, t_2+t_3, times velocity v_0\cos\theta. Thus,
e=\frac{R}{v_0\cos\theta \times (t_2 + t_3)} \\ = \frac{R}{v_0\cos\theta (2t' - t_1)} \\ = \frac{R}{\left( \frac{2v_0 \sin\theta}{g}-\frac{R}{v_0 \cos\theta}\right) v_0 \cos\theta} \\ = \frac{R}{\frac{2v_0^2 \sin\theta\cos\theta}{g}-R} \\ = \frac{R}{\frac{2v_0^2 \sin\theta\cos\theta-gR}{g}} \\ = \frac{gR}{2v_0^2 \sin\theta\cos\theta-gR}
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