A projectile is shot from ground level with an initial velocity $v_0$ at an angle $\theta $. It then barely clears the edge of a cliff of height $H=49.28\ m$ located a distance $L=40\ m$ from the launch point at $t_1=2\ s$. You may use $g\approx 10\ m/s^2$ in your calculations.

1. What is the initial speed $v_0$ of the projectile?

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The ball is hit from an initial position $\left\{ \begin{array}{c}
x_0=0 \
y_0=0 \end{array}
\right.$ and the components of the initial velocity ${\overrightarrow{v}}_0$ and of the acceleration due to gravity are

\left\{ \begin{array}{c} v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ } \\ v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} a_x=0 \ \ \ \ \\ a_y=-g \end{array} \right.

The kinematic equations describing the flight of the ball are therefore given by

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

However, since the initial speed $v_0$ and the launch angle $\theta $ are unknown, it is best to express these kinematic equations in terms of $v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ }$ rather than in terms of $v_0$ and $\theta $ (mostly because $v_0$ and $\theta $ appears in both $x\left(t\right)$ and $y\left(t\right)$ which makes it harder to solve for them because the equations are coupled). The kinematic equations expressed in terms of $v_{0x}$ and $v_{0y}$ are as follows

\left\{ \begin{array}{c} x\left(t\right)=v_{0x}t \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_{0y}t} \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} v_x\left(t\right)=v_{0x} \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ v_y\left(t\right)=-gt+v_{0y} \end{array} \right.

Solving for $v_{0x}$:
At time $t_1$, the projectile barely clears the edge of the cliff and we have $x\left(t_1\right)=L$. This allows us to solve for $v_{0x}$ as follows.

\begin{aligned} x\left(t_1\right)=2H\ \ \ \ &\Rightarrow \ \ \ \ \ v_{0x}t_1=L \\ \\ &\Rightarrow \ \ \ \ \ v_{0x}=\frac{L}{t_1} \\ \\ &\Rightarrow \ \ \ \ \ v_{0x}=\frac{40}{2}=20\ m/s \end{aligned}

Solving for $v_{0y}$:
At time $t_1$, the projectile barely clears the edge of the cliff and we have $y\left(t_1\right)=H$. This allows us to solve for $v_{0y}$ as follows.

\begin{aligned} y\left(t_1\right)=H\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_1+v_{0y}t_1=H \\ \\ &\Rightarrow \ \ \ \ \ v_{0y}t_1=H+\frac{1}{2}gt^2_1 \\ \\ &\Rightarrow \ \ \ \ \ v_{0y}=\frac{\displaystyle{H+\frac{1}{2}gt^2_1}}{t_1} \\ \\ &\Rightarrow \ \ \ \ \ v_{0y}=\frac{\displaystyle{49.28+\frac{1}{2}\cdot 10\cdot 2^2}}{2}=34.64\ m/s \end{aligned}

Computing the initial speed $v_0$:
Given the components $v_{0x}$ and $v_{0y}$, we conclude that the initial speed of the projectile is equal to

\boxed{v_0=\sqrt{v^2_{0x}+v^2_{0y}}=\sqrt{{20}^2+{34.64}^2}=40\ \ m/s}

2. What is the launch angle $\theta $?

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The launch angle $\theta $ can be derived from the components $v_{0x}$ and $v_{0y}$ as follows

{\mathrm{tan} \left(\theta \right)\ }=\frac{v_{0y}}{v_{0x}}\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{\theta ={{\mathrm{tan}}^{-1} \left(\frac{v_{0y}}{v_{0x}}\right)\ }={{\mathrm{tan}}^{-1} \left(\frac{34.64}{20}\right)\ }\approx 60{}^\circ}

3. How far past the edge of the cliff does the projectile land (on the horizontal plateau)?

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To derive the distance $d$ between the edge of the cliff and the point where the projectile lands, we first solve $y\left(t\right)=H$ as follows

\begin{aligned} y\left(t\right)=H\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2+v_{0y}t=H \\ \\ &\Rightarrow \ \ \ \ \ -\frac{1}{2}\cdot 10t^2+34.64t=49.28 \\ \\ &\Rightarrow \ \ \ \ \ 10t^2-69.28t+98.56=0 \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{t_1=\frac{69.28-\sqrt{{69.28}^2-4\cdot 10\cdot 98.56}}{20}=2.0\ s} \\ \\ \displaystyle{t_2=\frac{69.28+\sqrt{{69.28}^2-4\cdot 10\cdot 98.56}}{20}=4.9\ s} \end{array} \right. \end{aligned}

Thus, we conclude that the projectile clears the cliff at time $t_1=2\ s$ and lands at time $t_2=4.9\ s$. The $x$-location of the projectile when it lands is equal to $x\left(t_2\right)$ and therefore we conclude that the distance from the cliff when the projectile lands is equal to

\boxed{d=x\left(t_2\right)-L=v_{0x}t_2-L=20\cdot 4.9-40=58\ m}

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