A stone is dropped into water from a bridge 44.1 m above the water. Anoth

Discussion Forum : Motion In One Dimension

Question -

A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone?

Options:

A . 12.25 m/s

B . 14.75 m/s

C . 16.23 m/s

D . 17.15 m/s

Answer: Option A
:
A

Time taken by first stone to reach the water surface from the bridge be t, then

$h = u t + \frac{1}{2} g t^{2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8 t^{2}$

$t = \sqrt{\frac{2 \times 44.1}{9.8}} = 3 s e c$

Second stone is thrown 1 sec later and both strike simultaneously.This means that the time of travel for the second stone = 3 - 1 = 2 sec

Hence, 44.1 = $2 u + \frac{1}{2} 9.8 (2)^{2}$

$\Rightarrow 44.1 - 19.6 = 2 u \Rightarrow u = 12.25 m / s$

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