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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

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Solution

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 strikes simultaneously. This means that the time left for second stone = 3 - 1 = 2 sec
Hence 44.1=u×2+129.8(2)2
44.119.6=2uu=12.25ms

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