Ncert Exercises & Solutions

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

Let the speeds of the two balls (1and 2) be $v_{1}$ and $v_{2}$ where

$if v_{1} = 2 v, v_{2} = v$

If $y_{1} and y_{2}$ and the distance covered by

the balls 1 and 2, respectively, before coming to rest, then

$\begin{array}{l} y_{1} = \frac{v_{1}^{2}}{2 g} = \frac{4 v^{2}}{2 g} and y_{2} = \frac{v_{2}^{2}}{2 g} = \frac{v^{2}}{2 g} \\ Since, y_{1} - y_{2} = 15 m, \frac{4 v^{2}}{2 g} - \frac{v^{2}}{2 g} = 15 m or \frac{3 v^{2}}{2 g} = 15 m \\ or v^{2} = \sqrt{5 m \times (2 \times 10)} m / s^{2} \\ or v = 10 m / s \end{array}$

Clearly, $v_{1}$ = 20 m/s and $v_{2}$ = 10m/s

$\begin{array}{l} as y_{1} = \frac{v_{1}^{2}}{2 g} = \frac{(20 m)^{2}}{2 \times 10 m 15} = 20 m \\ y_{2} = y_{1} - 15 m = 5 m \end{array}$

If $t_{1}$ is the time taken by the ball 2 toner

$\begin{array}{l} a distance of 5 m then from y_{2} = v_{2} t - \frac{1}{2} g t_{2}^{2} \\ \begin{matrix} 5 = 10 t_{2} - 5 t_{2}^{2} or t_{2}^{2} - 2 t_{2} + 1 = 0 \\ {where t}_{2} = 15 \end{matrix} \end{array}$

Since, (time is taken by bail 1 to cover the distance of 20m) is 2s, the time interval between the two throws

$= t_{1} - t_{2} = 2 s - 1 s = 1 s$

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