Question

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

Answer 1

The ball's horizontal position in the air is

$x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t$

It hits the wall when $x=17.5\,\mathrm m$, which happens at

$17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s$

Meanwhile, the ball's vertical position is

$y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2$

where $g$ is the acceleration due to gravity, 9.80 m/s^2.

At the time the ball hits the wall, its vertical position (relative to its initial position) is

$y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}$