Question

A ball is thrown vertically into the air at $36m/s$. After 3 sec., another ball is thrown vertically, the initial velocity the second ball have to pass the first ball must at $30$ meter from the ground in (m/s) is___

• A. 25

Answer 1

The correct option is A 25


$1^{st}\text{Ball}$,
$S=ut-\frac{1}{2}g{t}^2$

$30=36t-4.905t^2$

$\therefore 4.905t^2-36t+30=0$

Solving above quadratic equation we get

$t=6.3809\text{sec}tot=0.9585\text{sec}$

At t=0.9585 sec the ball is going in upward direction and at t=6.3809sec the ball is going in downward direction after coming back from the highest point.

Clearly at $t=0.9585$ it is not possible to meet the two balls because second ball is fired at $3sec$

$\therefore 2^{nd}\text{ball}S=ut-\frac{1}{2}g{t}^2$

$\text{here time}t=t-3$

$\therefore 30=u(t-3)-4.905(t-3)62$

$t=6.38s$

$V=25.4567\frac{m}{s}$