A rock is thrown by a person at an angle of ${\theta }_0=45{}^\circ $. A cliff of height $H=150\ m$ is located a distance $L=30\ m$ away from the person.

1. What throwing speed $v_0$ is necessary to barely throw the rock over the cliff?

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We begin by setting up the kinematic equations of the rock after it is thrown. The acceleration of the rock while it is airborne is equal to the acceleration due to gravity and, therefore, we have $a_x=0$ and $a_y=-g$.

The initial velocity ${\overrightarrow{v}}_0$ has two components that are given by $\displaystyle{\left\{ \begin{array}{c} v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ } \\ v_{0y}=v_0{\mathrm{sin} \left({\theta }_0\right)\ } \end{array}
\right.}$ and we set the origin at the launching point. We then conclude that the kinematic equations describing the flight of the rock are given by

\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\left(t\right)=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.

In order for the rock to barely clear the top of the cliff, its speed $v_0$ must be such that when $x=L$ the height of the rock is $y=0$.

It takes the rock a time $t_1$ to reach $y=0$ and we solve for $t_1$ as follows

\begin{aligned} y\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_1=0 \\ \\ &\Rightarrow \ \ \ \ \ \left(v_0{\mathrm{sin} \left({\theta }_0\right)\ }-\frac{1}{2}gt_1\right)\cdot t_1=0 \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} t_1=0 \\ \\ or \\ \\ \displaystyle{t_1=\frac{2v_0{\mathrm{sin} \left({\theta }_0\right)\ }}{g}} \end{array} \right. \end{aligned}

Since $t_1=0$ is the time at which the rock is thrown, we disregard this solution. Thus, the time we are looking for is given by

t_1=\frac{2v_0{\mathrm{sin} \left({\theta }_0\right)\ }}{g}

For the position of the rock to satisfy $x\left(t_1\right)=L$, the launch speed $v_0$ must be such that

\begin{aligned} x\left(t_1\right)=L\ \ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_1=L \\ \\ &\Rightarrow \ \ \ \ \ \frac{2v^2_0{\mathrm{sin} \left({\theta }_0\right)\ }{\mathrm{cos} \left({\theta }_0\right)\ }}{g}=L \\ \\ &\Rightarrow \ \ \ \ \ v_0=\sqrt{\frac{gL}{2{\mathrm{sin} \left({\theta }_0\right)\ }{\mathrm{cos} \left({\theta }_0\right)\ }}} \end{aligned}

Therefore, the initial speed $v_0$ must be equal to

\boxed{v_0=\sqrt{\frac{gL}{2{\mathrm{sin} \left({\theta }_0\right)\ }{\mathrm{cos} \left({\theta }_0\right)\ }}}=\sqrt{\frac{9.8\cdot30}{2{\mathrm{sin} \left(45\right)\ }{\mathrm{cos} \left(45\right)\ }}}\approx 17.14\ m/s}

2. Assuming that the rock barely goes over the cliff, at what horizontal distance from the base of the cliff does it hit the ground?

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If the rock barely goes over the cliff, then its speed is equal to the value $v_0$ found in question 1. In addition, the angle of the rock when it reaches $x=L$ is equal to ${\theta }_1={\theta }_0=45{}^\circ $ with respect to the horizontal (angled below the horizontal). Therefore, we can now setup the different kinematic equations for the rock for the second part of the fall.

The acceleration of the rock while it is airborne is equal to the acceleration due to gravity and, therefore, we have $a_x=0$ and $a_y=-g$.

The initial velocity ${\overrightarrow{v}}_1$ has two components that are given by $\displaystyle{\left\{ \begin{array}{c} v_{1x}=v_1{\mathrm{cos} \left({\theta }_1\right)\ } \\ v_{1y}=-v_1{\mathrm{sin} \left({\theta }_1\right)\ } \end{array} \right.}$ where $v_{1y}$ is negative because the rock is moving downward initially when it reaches the edge of the cliff. Setting the origin at the base of the cliff for this phase of the motion, we derive that the kinematic equations of the rock are

\left\{ \begin{array}{c} x\left(t\right)=v_1{\mathrm{cos} \left({\theta }_1\right)\ }t \\ \\ v_x\left(t\right)=v_1{\mathrm{cos} \left({\theta }_1\right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2-v_1{\mathrm{sin} \left({\theta }_1\right)\ }t+H} \\ \\ v_y\left(t\right)=-gt-v_1{\mathrm{sin} \left({\theta }_1\right)\ } \end{array} \right.

The rock hits the ground at a time $t_2$ such that $y\left(t_2\right)=-H$ and we solve for $t_2$ as follows

\begin{aligned} y\left(t_2\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_2-v_1{\mathrm{sin} \left({\theta }_1\right)\ }t_2+H=0 \\ \\ &\Rightarrow \ \ \ \ \ 4.9t^2_2+12.12t_2-150=0 \end{aligned}

Solving the quadratic equation yields two answers

\left\{ \begin{array}{c} t_2\approx -6.90\ s \\ \\ or \\ \\ t_2\approx 4.43\ s \end{array} \right.

Keeping only the positive value, we find the horizontal distance covered by the rock during the second part of the motion is equal to

\boxed{x\left(t_2\right)=v_1{\mathrm{cos} \left({\theta }_1\right)\ }t_2\approx 17.14\cdot {\mathrm{cos} \left(45\right)\ }\cdot 4.43\approx 53.17\ m}

Thus, the rock hits the ground at a distance $53.17\ m$ from the base of the cliff.

3. What is the speed of the rock when it hits the ground?

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The rock reaches the ground at time $t_2\approx 4.43\ s$ at which the components of its velocity are equal to

\begin{aligned} v_x\left(t_2\right)&=v_1{\mathrm{cos} \left({\theta }_1\right)\ }\approx 17.14\cdot {\mathrm{cos} \left(45\right)\ }\approx 12.12\ m/s \\ \\ v_y\left(t_2\right)&=-gt_2-v_1{\mathrm{sin} \left({\theta }_1\right)\ }\approx -9.8\cdot 4.43-17.14\cdot {\mathrm{sin} \left(45\right)\ }\approx -55.53\ m/s \end{aligned}

Thus, the speed of the rock upon impact is equal to

\boxed{v=\sqrt{v_x{\left(t_2\right)}^2+v_y{\left(t_2\right)}^2}=\sqrt{{12.12}^2+{55.53}^2}\approx 56.83\ m/s}

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