Question

A ball is dropped from a high rise platform at $t=0$ starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=18s$. What is the value of $v$? (Take $g=10m/s^2$)

Answer 1

The distance traveled by the first ball in 18s is $h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times (18{)}^2=1620m$

Now to meet second ball has to same distance in$(18-6)=12s$

So for second ball,

$h=vt+\frac{1}{2}g{t}^2$

$1620=v\times 12+\frac{1}{2}\times 10\times (12{)}^2$

$v=\frac{1620-720}{12}=75m/s$