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 $t=18\text{s}$. What is the value of $v$ ?
(take $g=10{\text{m/s}}^2$)

• A. $75\text{m/s}$
• B. $55\text{m/s}$
• C. $60\text{m/s}$
• D. $40\text{m/s}$

Answer 1

The correct option is A $75\text{m/s}$

Let the two balls meet at depth $h$ from platform
So, comparing both the cases, we have
$h=\frac{1}{2}g(18{)}^2=v(12)+\frac{1}{2}g(12{)}^2$
$\Rightarrow 12v=\frac{1}{2}g({18}^2-{12}^2)=900$
$\Rightarrow v=75{\text{ms}}^{-1}$.