P02S2013 – Throwing a Rockk Over a Plateau

The kinematic equations for the first part of the motion of the rock (until it reaches the plateau) are:

\left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t} \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
v_x(t)=v_0{\mathrm{cos} \left({\theta }_0\right)\ } \\
\\
v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left({\theta }_0\right)\ } \end{array}
\right.

If the rock just reaches the top of the plateau, then it reaches its maximum height just when it reaches the plateau. At that point, its velocity is purely horizontal $\left(v_y=0\right)$ which explains why its slides along the plateau instead of flying past it or bouncing on it. Thus, since the horizontal velocity during the flight of the rock is constant and equal to $v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }$, we conclude that the speed $v_{top}$ of the rock when it reaches the plateau is $v_{top}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }$. In addition, since the plateau is frictionless, the speed of the rock remains constant and equal to $v_{top}$ as the rock slides on the plateau.

This speed $v_{top}$ must be such that the rock travels a distance $D=4\cdot 9.8\ m$ in a time $T=2.0\ s$ and is therefore equal to

v_{top}=\frac{D}{T}=\frac{4\cdot 9.8}{2}=19.6\ m/s

Thus, we conclude that the $x$-component $v_{0x}$ of the initial velocity of the rock is equal to

\boxed{v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }=19.6\ m/s}

To derive the launch angle ${\theta }_0$, we need to determine the $y$-component $v_{0y}$ of the initial velocity of the rock. Because we know the rock reaches the left edge of the plateau when it is at its maximum height $\left(v_y=0\right)$, conclude that the $y$-component of the initial velocity $v_{0y}=v_0{\mathrm{sin} \left({\theta }_0\right)\ }$ is such that the acceleration due to gravity $g$ cancels $v_{0y}$ as the rock rises to the height $H$ of the plateau.

Thus, we solve for $v_{0y}$ as follows

\begin{aligned}
v^2_{fy}=v^2_{iy}+2a_y\mathrm{\Delta }y\ \ \ \ \ &\Rightarrow \ \ \ \ \ 0^2=v^2_{0y}-2gH \\
\\
&\Rightarrow \ \ \ \ \ v_{0y}=\sqrt{2gH} \\
\\
&\Rightarrow \ \ \ \ \ v_{0y}=\sqrt{4g^2}=2\cdot 9.8=19.6\ m/s
\end{aligned}

The components of the initial velocity of the rock are therefore equal to

\begin{aligned}
v_{0x}&=19.6\ m/s \\
\\
v_{0y}&=19.6\ m/s
\end{aligned}

Since the components are equal, we conclude that the launch angle is equal to

\boxed{{\theta }_0={{\mathrm{tan}}^{-1} \left(\frac{v_{0y}}{v_{0x}}\right)\ }={{\mathrm{tan}}^{-1} \left(\frac{19.6}{19.6}\right)\ }=45{}^\circ}

By symmetry, if $t_1$ denotes the time it takes the rock to reach the left end of the plateau, then it will take the rock a time $t_1$ after leaving the right end of the plateau to hit the ground. This is because the rock lands on the left end with horizontal velocity $v_{0x}$ that remains constant (no friction) as it slides across the plateau and flies off the right end.

Thus, we first solve for $t_1$ by arguing that it satisfies $v_y\left(t_1\right)=0$ as follows

\begin{aligned}
v_y\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ -gt_1+v_0{\mathrm{sin} \left({\theta }_0\right)\ }=0 \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{v_0{\mathrm{sin} \left({\theta }_0\right)\ }}{g} \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{19.6}{9.8}=2\ s
\end{aligned}

In the time $t_1$, the rock travels a distance $x\left(t_1\right)$ between its launch point to the left edge that is equal to

x\left(t_1\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_1=19.6\cdot 2=39.2\ m

It then travels a distance $D$ across the plateau, followed by another amount $x\left(t_1\right)$ as it flies off the right edge and hits the ground. The total distance traveled by the rock is therefore equal to

\boxed{D_{tot}=2x\left(t_1\right)+D=2\cdot 39.2+4\cdot 9.8=117.6\ m}