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

Answer: (A)

20 m s^-1, 10 m s^-1

Answer 2

For first stone. Taking the vertical upwards motions of the first stone up to highest point

Here , $u = u_{1},v = 0$ (At highest point velocity is zero )

个

$a = - g,S = h_{1}$

$\therefore(0)^{2} - u_{1}^{2} = 2( - g)h_{1}$

Or $h_{1} = \frac{u_{1}^{2}}{2g}$ …… (1)

For second stone. Taking the vertical upwards motions of the second stone up to highest point

Here, $u = u_{2},v = 0,a = - g,S = h_{2}$

As $v^{2} - u^{2} = 2as$

$\therefore(0)^{2} - u_{2}^{2}2( - g)h_{2}$

$h_{2} = \frac{u_{2}^{2}}{2g}$ ……… (ii)

As per questions

$h_{1} - h_{2} = 15m,u_{2} = \frac{u_{1}}{2}$

Subtract (i) From (ii), 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 $u_{1} = 20ms^{- 1}$and $u_{2} = \frac{u_{1}}{2} = 10ms^{- 1}$