Bill and Ted are on the roofs of two buildings separated by a distance $D=100\ m$ (see figure). The height of both buildings is $H=19.6\ m$. One hundred meters is farther than Bill can throw his superball, so he decides to throw the ball so that it bounces once on the ground between the two buildings before it reaches Ted. Bills throws the ball so that when it leaves his hand, its velocity is parallel to the ground. We want to find the minimum velocity $v_0$ with which the ball should be thrown so that it just reaches Ted. You should know that the superball is such that when it hits the ground:

\begin{aligned} v_x\ \left(before\ hitting\right)&=v_x\left(after\ hitting\right) \\ \\ v_y\left(before\ hitting\right)&=-v_y(after\ hitting) \end{aligned}

1. Draw the above figure and sketch what you think the trajectory of the ball should be.

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The barely reaches Ted and therefore arrives at Ted’s location with a horizontal velocity. Indeed, if it had upward velocity at that point, it would still move upward and would not “barely reach” Ted. If it had a downward velocity, it simply would not reach Ted because the ball cannot reach a height greater than $H$ as it was thrown by Bill horizontally from a height $H$.

When the superball bounces off the ground, the $x$-component $v_x$ of its velocity before impact remains the same, which is what $v_x\ \left(before\ hitting\right)=v_x\left(after\ hitting\right)$ is telling you. This means that in the time it takes the ball to hit the ground, its velocity $v_x$ (which is constant) is such that the ball travels a distance $D/2$. Thus, in the time it takes the ball to rise back up to Ted’s location, it travels another distance $D/2$.

When the superball bounces off the ground, the $y$-component $v_y$ of its velocity before impact is flipped, which is what $v_y\left(before\ hitting\right)=-v_y(after\ hitting)$ is telling you. This means that the ball bounces up with exactly the same vertical speed it hit the ground with. Thus, it has enough vertical speed to rise back up to the height $H$ exactly by the time it reaches Ted’s location.

2. What is $v_0$?

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For the first half of its motion, the acceleration of the ball has components

\left\{ \begin{array}{c} a_x&=0 \ \ \ \\ a_y&=-g \end{array} \right.

and its initial velocity is horizontal with components

\left\{ \begin{array}{c} v_{0x}=v_0 \\ v_{0y}=0 \ \ \end{array} \right.

The kinematic equations giving the coordinates $x\left(t\right)$ and $y\left(t\right)$ of the ball throughout its fall are therefore given by

\left\{ \begin{array}{c} \displaystyle{x\left(t\right)=\frac{1}{2}a_xt^2+v_{0x}t+x_0=v_0t} \\ \\ \displaystyle{y\left(t\right)=\frac{1}{2}a_yt^2+v_{0y}t+y_0=H-\frac{1}{2}gt^2} \end{array} \right.

The time $t_1$ at which the ball hits the ground satisfies $\displaystyle{\left\{ \begin{array}{c}
x\left(t_1\right)=D/2 \
y\left(t_1\right)=0 \end{array}
\right.}$ and we solve for $t_1$ and $v_0$ as follows

\begin{aligned} \left\{ \begin{array}{c} \displaystyle{x\left(t_1\right)=\frac{D}{2}} \\ \\ y\left(t_1\right)=0 \end{array} \right.\ \ \ \ &\Rightarrow \ \ \ \ \left\{ \begin{array}{c} \displaystyle{v_0t_1=\frac{D}{2}} \\ \\ \displaystyle{-\frac{1}{2}gt^2_1+H=0} \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{v_0=\frac{D}{2t_1}} \\ \\ \displaystyle{t^2_1=\frac{2H}{g}} \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{v_0=\frac{D}{2t_1}=\frac{D}{2}\sqrt{\frac{g}{2H}}} \\ \\ \displaystyle{t_1=\sqrt{\frac{2H}{g}}} \end{array} \right. \end{aligned}

Thus, the speed $v_0$ at which Bill throws the superball is equal to

\boxed{v_0=\frac{D}{2}\sqrt{\frac{g}{2H}}=\frac{100}{2}\sqrt{\frac{9.8}{2\cdot 19.6}}=\frac{100}{4}=25\ m/s}

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