Question

If a ball is thrown straight up off the roof of a 40 meter high building at 20m/s, how long must the thrower wait to drop a second ball from rest at the top of the building so that both balls strike the ground at the same time?

Answer 1

$s=gt^2/2\to t=\sqrt{2s/g}=\sqrt{2\cdot40/9.8}=2.86\ (s)$

$v=v_0-gt_1\to t_1=(v_0-v)/g=(20-0)/9.8=2.04 \ (s)$

$s_1=\frac{v^2-v_0^2}{2g}=\frac{0-20^2}{2\cdot(-9.8)}=20.41\ (m)$

$s_0+s_1=gt_2^2/2\to t_2=\sqrt{2(s_0+s_1)/g}=\sqrt{2(40+20.41)/9.8}=3.51\ (s)$

$t_3=3.51-2.86=0.65\ (s)$

$\Delta t=t_1+t_3=2.04+0.65=2.69\ (s)$ . Answer