Question

A man is standing on top of a building $100\, m$ high. He throws two balls vertically upwards, one at $t = 0$ and other after a time interval (less than $2 \,s$). 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. The velocities with which the balls were thrown are (Take $g= 10 \,m\, s^{-2}$)

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

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 $v^{2}-u^{2}=2aS$
$\therefore \left(0\right)^{2}-u^{2}_{1}=2\left(-g\right)h_{1}$ or $h_{1}=\frac{u^{2}_{1}}{2g}\quad\ldots\left(i\right)$
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 $h_{2}=\frac{u^{2}_{2}}{2g}\quad\ldots\left(ii\right)$
As per question
$h_{1}-h_{2}=15\,m, u_{2}=\frac{u_{1}}{2}$
Subtract $\left(ii\right)$ from $\left(i\right)$, we get ;
$h_{1}-h_{2}=\frac{u^{2}_{1}}{2g}-\frac{u^{2}_{2}}{2g}$
On substituting the given information, we get
$15=\frac{u^{2}_{1}}{2g}-\frac{u^{2}_{1}}{8g}=\frac{3u^{2}_{1}}{8g}$ or $u^{2}_{1}=\frac{15\times8g}{3}$
$=\frac{15\times8\times10}{3}=400$
or $u_{1}=20\,m\,s^{-1}$ and $u_{2}=\frac{u_{1}}{2}=10\,m\,s^{-1}$