Question

A ball is dropped from a bridge $122.5\text{m}$ high. After the first ball has fallen for $2$ second, a second ball is thrown straight down after it, what must be the initial velocity of the second ball, so that both the balls hit the surface of water at the same time?

• A. $26.1\text{m/s}$
• B. $9.8\text{m/s}$
• C. $55.5\text{m/s}$
• D. $49\text{m/s}$

Answer 1

The correct option is A $26.1\text{m/s}$

Time taken by the first object to reach the ground $=t$, so
$122.5=ut+\frac{1}{2}g{t}^2$
$122.5=\frac{1}{2}\times 10\times t^2$
$\Rightarrow t=5$ sec(approx)
Time to be taken by the second ball to reach the ground
$=(5-2)\text{sec}=3\text{sec}$.

If $u$ be its initial velocity then,
$122.5=u\times 3+\frac{1}{2}g{t}^2=3u+\frac{1}{2}\times 10\times 9$
$\Rightarrow 3u=122.5-45=77.5$
$\therefore u=26$ (approx).