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Question

# A stone is dropped into water from a bridge 44.1m 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

A
12.25 m/s
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B
14.75 m/s
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C
16.23 m/s
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D
17.15 m/s
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Solution

## The correct option is A 12.25 m/sTime taken by first stone to reach the water surface from the bridge be t, then h=ut+12gt2⇒44.1=0×t+12×9.8t2 t=√2×44.19.8=3 sec Second stone is thrown 1 sec later and both strikes simultaneously.This means that the time left for second stone=3-1 = 2 sec. Hence 44.1=u×2+129.8(2)2 ⇒44.1−19.6=2u⇒u=12.25m/s

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