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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?


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/s


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

h=ut+12gt244.1=0×t+12×9.8t2

t=2×44.19.8=3sec

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 = 2u+129.8(2)2

44.119.6=2uu=12.25m/s


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