Projectile onto an inclined plane: The maximum range

Question:
A projectile is launched from the origin with initial velocity $v_0$ and the angle $\theta$ from the $x$-axis. The object is reached on the inclined plane below $x$-axis with angle of $\alpha$. Assume that the mass, initial velocity and angle $\alpha$ are fixed. Then, what is the angle $\theta$ to obtain maximum range on the inclined plane?

Answer:
The position of the projectile is given as
\begin{eqnarray}
x &=& v_0\cos\theta \ t  \\
y &=& v_0\sin\theta \ t - \frac{1}{2}gt^2
\end{eqnarray}
For the inclined plane, we can have the relationship between $x$ and $y$ as follows:
\begin{equation}
y = -x\tan\alpha
\end{equation}
Use (3) to eliminate $y$.
\begin{equation}
-x\tan\alpha = v_0 \sin\theta \ t - \frac{1}{2}gt^2
\end{equation}
From (1), we have $t = \frac{x}{v_0 \cos\theta}$ to eliminate $t$.
\begin{eqnarray}
-x\tan\alpha = v_0 \sin\theta\frac{x}{v_0 \cos\theta} - \frac{1}{2}g\left(\frac{x}{v_0 \cos\theta}\right)^2   \\
-x\tan\alpha = x\tan\theta - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} \\
x(\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} = 0  \\
x \left\{ (\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx}{v_0^2 \cos^2\theta} \right\} = 0
\end{eqnarray}
$x=0$ is also the solution, but we are interested in the other one. Now, we can solve for $x$ as follows:
\begin{eqnarray}
x &=& \frac{2v_0^2\cos^2\theta (\tan\alpha + \tan\theta)}{g}\\
  &=& \frac{2v_0^2\cos^2\theta \left(\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\theta}{\cos\theta} \right)}{g}\frac{\cos\alpha}{\cos\alpha}\\
 &=& \frac{2v_0^2\cos\theta (\sin\alpha\cos\theta + \sin\theta\cos\alpha)}{g\cos\alpha} \\
 &=& \frac{2v_0^2\cos\theta \sin(\alpha+\theta)}{g\cos\alpha} \\
 &=& \frac{2v_0^2 \left\{ \frac{1}{2}\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha} \\
 &=& \frac{v_0^2 \left\{\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha}
\end{eqnarray}
$v_0$ and $\alpha$ are constant, so the value of $\sin(2\theta+\alpha)$ affects the value of the whole function. A sine or cosine function oscillates between -1 and 1 assuming that the amplitude is 1. Thus, when we have
\begin{equation}
\sin(2\theta+\alpha) = 1
\end{equation}
the range becomes maximum. The angle must be
\begin{eqnarray}
2\theta+\alpha &=& \sin^{-1}1 = \frac{\pi}{2} \\
2\theta &=& \frac{\pi}{2}-\alpha  \\
\theta &=& \frac{\pi}{4} - \frac{\alpha}{2}
\end{eqnarray}

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A projectile motion: Bouncing back from a wall

Question:
As shown in the figure, an object is launched at initial velocity $v_0$ with the angle of $\theta$ toward the wall. The distance between A and B is $R$. The wall is frictionless and the coefficient of restitution with the object is $e$. The initially launched object bounces at C and comes back to the original position A through the projectile D. Assume that there is no air resistance. Answer the following questions:
(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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