Question

A man is standing on top of a building $100m$ high. He throws two balls vertically upwards, one at $t=0$ and other after a time interval (less than $2s$). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $15m$ at $t=2s$. The gap is found to remain constant. The velocities with which the balls were thrown are (Take $g=10m{s}^{-2}$)

• A. $20m{s}^{-1}$, $10m{s}^{-1}$
• B. $10m{s}^{-1}$, $5m{s}^{-1}$
• C. $16m{s}^{-1}$, $8m{s}^{-1}$
• D. $30m{s}^{-1}$, $15m{s}^{-1}$

Answer 1

Answer: (A)

For first stone.
Taking the vertical upwards motion of the first stone up to highest point
Here, $u-u_1$, $v=0$ (At highest point velocity is zero)
$a=-g$ , $S=h_1$
As $v^2-u^2=2aS$
$\therefore {\left(0\right)}^2-u_1^2=2\left(-g\right)h_1$ or $h_1=\frac{u_1^2}{2g}\dots \left(i\right)$
For second stone,
Taking the vertical upwards motion of the second stone up to highest point

Here, $u=u_2$, $v=0$, $a--g$, $S=h_2$
As $v^2-u^2=2aS$
$\therefore {\left(0\right)}^2-u_2^2=2-\left(-g\right)h_2$ or $h_2=\frac{u_2^2}{2g}\dots \left(ii\right)$
As per question
$h_1-h_2=15m,u_2=\frac{u_1}{2}$
Subtract $\left(ii\right)$ from $\left(i\right)$, we get ;
$h_1-h_2=\frac{u_1^2}{2g}-\frac{u_2^2}{2g}$
On substituting the given information, we get
$15=\frac{u_1^2}{2g}-\frac{u_1^2}{8g}=\frac{3u_1^2}{8g}$ or $u_1^2=\frac{15\times 8g}{3}$
$=\frac{15\times 8\times 10}{3}=400$
or $u_1=20m{s}^{-1}$ and $u_2=\frac{u_1}{2}=10m{s}^{-1}$