Question

A ball is thrown vertically upwards at 5 m/s from the roof of a 100 m building. A second ball is thrown downwards from the same point 2 seconds later at 20 m/s. a) Where and when will the balls meet? b) What are their velocities at that instant?

Answer 1

$t_1=\frac vg=0.5~s,$

$t=t_2-t_1=1.5~s,$

$s=\frac{v^2}{2g}=1.25~m,$

a)

$s'=s+h-\frac{gt'^2}2,$

$s'=h-v'(t'-t)-\frac{g(t'-t)^2}2,\implies$

$t'=\frac{v't-\frac{gt^2}2+s}{v'-gt}=4~s,$

$s'=21.25~m,$

b)

$v_{b1}=gt'=40~\frac ms,$

$v_{b2}=v'+g(t'-t)=65~\frac ms.$