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\,(\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=(90.0\,\mathrm m)/(v_(0x))`

Substituting this into the second equation gives

`6.50\,\mathrm m=v_(0y)\left((90.0\,\mathrm m)/(v_(0x))\right)-\frac g2\left((90.0\,\mathrm m)/(v_(0x))\right)^2`

`6.50\,\mathrm m=90\tan35^\circ-((90.0\,\mathrm m)^2g)/(2|v_0|^2\cos^235^\circ)`

`\implies|v_0|^2=1046.55\,(\mathrm m^2)/(\mathrm s^2)`

`\implies|v_0|=32.4\,(\mathrm m)/(\mathrm s)`