A man is standing on top of a building 100m high. He throws two balls vertically, one at $t = 0$ and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15$ m at $t = 2$ s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Let the speeds of the two balls (1 and 2) be $v_1$ and $v_2$ where
if $v_1 = 2v, v_2 = v$
If $y_1$ and $y_2$ and the distance covered by the balls 1 and 2, respectively, before coming to rest, then
$y_1 = \frac{v_1^2}{2g} = \frac{4v^2}{2g}$ and $y_2 = \frac{v_2^2}{2g} = \frac{v^2}{2g}$
$y_1 - y_2 = 15m$
$\frac{4v^2}{2g} - \frac{v^2}{2g} = 15m$ or $ v^2 = \frac{15m \times (2 \times 10)}{3} $
or $v = 10 \; m/s$
Clearly, $v_1 = 20 m/s$ and $v_2 = 10 m/s$
as $y_1 = \frac{v_1^2}{2g} = \frac{(20m)^2}{2 \times 10m/s^2} = 20m$
$y_2 = y_1 - 15m = 5m$
If $t_2$ is the time taken by the ball 2 to ner a distance of 5 m, then from $y_2 = v_2t - \frac{1}{2}gt^2$
$5 = 10t_2 - \frac{5t_2^2}{2}$ or $t_2^2 - 2t_2 + 1 = 0,$
where $t_2 = 1s$
Since $t_1 $ (time taken by ball 1 to cover distance of 20 m ) is 2s, time interval between the two thvnws
$ = t_1 - t_2 = 2s - 1s = 1s $
Note We should be very careful, when we are applying the equation of rectilinear motion.