Question

A man is standing on top of a building 100 m high. He trows two balls vertically, one at t = 0 and other after a time interval (less than 2 second). The later ball is thrown at a velocity of half the first. The vertical gap b/w first and second ball is +15 m at t = 2s. The gap is found to remain constant. Calculate the velocity with the balls were thrown and the exact time interval b/w their throw.

Answer 1

Let the first ball be thrown with the velocity $u$ and second ball with a time interval $x$. Taking vertical motion of ball first for time $(t+x)$, we have

                      $S_1=u(t+x)-\dfrac{1}{2}g(t+x)^2$                 ...($i$)

Taking upward motion of second ball for time $t$, we have

                      $S_2=\dfrac{u}{2}t-\dfrac{1}{2}gt^2$                 ...($ii$)

∴                     $S_1-S_2=u\Bigg(t+x-\dfrac{t}{2}\Bigg)-\dfrac{1}{2}g(\cancel{t^2}+2xt+x^2\cancel{-t^2})$

                              $=u\Bigg(\dfrac{t}{2}+x\Bigg)-\dfrac{1}{2}g(2xt+x^2)$

As per the question, $S_1-S_2=15\text{ m}$ ; $t=2\text{ s}$ and $x=1\text{ s}$   [as $x<2\text{ s}$]

Then,                $15=u\Bigg(\dfrac{\cancel{2}}{\cancel{2}}+1\Bigg)-\dfrac{1}{2}\times10(2\times1\times2+1^2)$

or                     $15=u\times2-25$

or                     $u=\dfrac{40}{2}=20\text{ m/s}$

The velocity of the other ball = $\dfrac{u}{2}$ = $\dfrac{20}{2}$ = 10 m/s. Time interval $x$ = 1 s