Question

A baseball player throws a ball straight up and catches it 3.0 s later. If the ball accelerates at 9.85 m/s2 down, at what velocity did he throw the ball?

Answer 1

With constant acceleration, we have

$a=a_{\mathrm{av}}=\dfrac{v-v_0}t$

At the end of the throw, the ball's velocity is equal to its initial velocity in magnitude, but in the downward direction so its sign is opposite and $v=-v_0$. Then

$-9.85\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{-2v_0}{3.0\,\mathrm s}\implies v_0=15\,\dfrac{\mathrm m}{\mathrm s}$