Question

A ball is dropped from a bridge 122.5 m above a river. After 2s, a second ball is thrown down after it. What must its initial velocity be so that both hit the water at the same time ?

• A. 49 m/s
• B. 55.5 m/s
• C. 26.1 m/s
• D. 9.8 m/s

Answer 1

The correct option is C 26.1 m/s

Time taken by $1^{st}$ ball, $t=\sqrt{\frac{2h}{g}}$
$t=\sqrt{\frac{2\times 122.5}{g}}=\sqrt{\frac{245}{9.8}}$
$t=\sqrt{\frac{245}{49}\times \frac{10}{2}}=5sec$

So, some distance has to be covered by second ball in:
$3(5-2)$seconds:$\Rightarrow 122.5=u\left(3\right)+\frac{1}{2}\times 9.8(3{)}^2$
$u=26.1m/s$