Question

A man is standing on top of a building 100 m high. He throws two balls vertically, 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

The correct option is A 20 m $s^{-1}$, 10 m $s^{-1}$

Given:

The height of the building is $=100m$

The first ball is thrown at the instant $t=0$ and another ball is thrown after a time interval of $x$.

Let, the initial velocity of first ball$=um/s$.

The distance covered by the ball is given by:

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

On the other hand, the second ball is thrown at half the speed as of the first ball. So, the distance covered by the second ball is given by:

$S_2=\frac{u}{2}t-\frac{1}{2}g{t}^2$

The distance between the two balls at $t=2s$ is $15m$.

$\therefore S_1-S_2=15$

Now, $S_1-S_2=u(t+x-\frac{t}{2})-\frac{1}{2}g(t^2+2xt+x^2-t^2)$

Substitute the values in above equation:

Consider the time interval $x=1$ since it is less than 2 sec.

$15=u(1+\frac{2}{2})-\frac{1}{2}\times 10(2\times 1\times 2+1)$

$u=\frac{40}{2}m/s=20m/s$

Therefore, velocity of first ball$=20m/s$ in upward direction

Velocity of second ball is $\frac{20}{2}=10m/s$ in upward direction.