A ball is droped from a high rise platform $t=0$ starting from rest. After $6s$ 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 $v$?

The correct option is A 75 m/s

The distance traveled by first ball in $t=18s$ is $h_1=ut+\frac{1}{2}g{t}^2=0\left(18\right)+0.5\left(10\right)(18{)}^2=1620$

As the second ball travels same distance $h_1$ in time $t=(18-6)=12s$, so

$1620=vt+0.5g{t}^2$ or $1620=12v+0.5\left(10\right)(12{)}^2$ or $12v=900$

So, $v=75m/s$