Question

A basketball player hits a baseball which just clears a 7.50 m high, located 90.0 m away from home plate. If the ball was hit 1.0 m of the ground at an angle 35 degree, what was the initial speed of the baseball?

Answer 1

Let $v_0$ be the initial velocity of the ball. Then it has components

$\begin{cases}v_{0x}=|v_0|\cos35^\circ\\v_{0y}=|v_0|\sin35^\circ\end{cases}$

The position vector $(x,y)$ of the ball at time $t$ has components

$\begin{cases}x=v_{0x}t\\y=1.0\,\mathrm m+v_{0y}t-\frac g2t^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. When the ball is 90.0 m away from home plate, we're told that it clears (presumably) a barrier of some kind that is 7.50 m tall, so that for some time $t$ we get

$\begin{cases}90.0\,\mathrm m=v_{0x}t\\7.50\,\mathrm m=1.0\,\mathrm m+v_{0y}t-\frac g2t^2\end{cases}$

Solving for this $t$ gives

$t=\dfrac{90.0\,\mathrm m}{v_{0x}}$

Substituting this into the second equation gives

$6.50\,\mathrm m=v_{0y}\left(\dfrac{90.0\,\mathrm m}{v_{0x}}\right)-\dfrac g2\left(\dfrac{90.0\,\mathrm m}{v_{0x}}\right)^2$

$6.50\,\mathrm m=90\tan35^\circ-\dfrac{(90.0\,\mathrm m)^2g}{2|v_0|^2\cos^235^\circ}$

$\implies|v_0|^2=1046.55\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies|v_0|=32.4\,\dfrac{\mathrm m}{\mathrm s}$