Question

A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 3.5s with what velocity did it leave the ground

Answer 1

The ball's velocity at time $t$ would be given by

$v_y=v_{0y}-gt$

At its highest point, we'll have $v_y=0$. We're told this occurs at $t=3.5\,\mathrm s$, which means

$0=v_{0y}-\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.5\,\mathrm s)$

$\implies v_{0y}=34\,\dfrac{\mathrm m}{\mathrm s}$