Question

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the instant the first ball is released. determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. (use h and g as appropriate in your equation.)

Answer 1

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$, is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$