A ball is dropped from rest at a height $h_0$ above the ground and another one, located below the first one along the same vertical direction, is simultaneously tossed vertically upward from the ground with an unknown initial speed $v_0$.

1. Assuming the two balls collide, determine the time $T$ at which the collision occurs in terms of $h_0$ and $v_0$.

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We label ball #1 the ball dropped from rest and ball #2 the ball tossed upward.

Ball 1 experiences an acceleration $a_y=-g$, has no initial velocity, and its initial height is $y_0=h$. Its height $y_1\left(t\right)$ is therefore given by

y_1\left(t\right)=-\frac{1}{2}gt^2+h_0

Ball 2 experiences an acceleration $a_y=-g$, has an upward initial velocity $v_{0y}=v_0$ and no initial height. Its height $y_2\left(t\right)$ is therefore given by

y_2\left(t\right)=-\frac{1}{2}gt^2+v_0t

If the two balls collide at time $t=T$, then $T$ satisfies $y_1\left(T\right)=y_2\left(T\right)$ and is given by

\begin{aligned} y_1\left(T\right)=y_2\left(T\right)\ \ \ \ \ &\Rightarrow \ \ \ \ \ h_0=v_0T \\ \\ &\Rightarrow \ \ \ \ \ T=\frac{h_0}{v_0} \end{aligned}

The two balls collide at time $T$ equal to

\boxed{T=\frac{h_0}{v_0}}

2. Under the same assumption, determine the height $H$ at which the two balls collide, in terms of $g$, $h_0$, and $v_0$.

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The height $H$ at which the two balls collide is equal to their height at time $t=T$ and we write

\boxed{H=y_1\left(T\right)=-\frac{1}{2}gT^2+h_0=-\frac{gh^2_0}{2v^2_0}+h_0}

3. On two separate graphs, plot the position vs. time of both balls for the scenario when the two balls collide in the air and the scenario when the two balls never collide.

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The two balls will collide in the air if the initial speed $v_0$ with which the second ball is tossed upward is large enough. Indeed, if $v_0$ is too small, the second ball will rise and fall in a shorter time than the time it takes the first ball to fall and collide with it.

4. Calculate the minimum initial speed of the tossed ball that allows a collision in the air.

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The last point where the two balls can collide in the air is if they collide at ground level. If $t_f$ is the time it takes for the ball to reach the ground (at the same time in this case), then their positions must satisfy $y_1\left(t_f\right)=y_2\left(t_f\right)=0$. This occurs when the initial speed of the tossed ball is the smallest it can be which is when $v_0=v_{min}$.

Thus, we first determine $t_f$ by solving $y_1\left(t_f\right)=0$ as follows

\begin{aligned} y_1\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_f+h_0=0 \\ \\ &\Rightarrow \ \ \ \ \ t_f=\sqrt{\frac{2h_0}{g}} \end{aligned}

Then, we solve $y_1\left(t_f\right)=y_2\left(t_f\right)$ to derive $v_{min}$

\begin{aligned} y_1\left(t_f\right)=y_2\left(t_f\right)\ \ \ \ &\Rightarrow \ \ \ -\frac{1}{2}gt^2_f+h_0=-\frac{1}{2}gt^2_f+v_{min}t_f \\ \\ &\Rightarrow \ \ \ \ \ h_0=v_{min}t_f \\ \\ &\Rightarrow \ \ \ \ \ v_{min}=\frac{h_0}{t_f}=h_0\sqrt{\frac{g}{2h_0}}=\sqrt{\frac{gh_0}{2}} \end{aligned}

Thus, the minimum initial speed of the tossed ball that allows a collision in the air is equal to

\boxed{v_{min}=\sqrt{\frac{gh_0}{2}}}

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