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?
Answer: Option A
:
A
Time 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=3sec
Second stone is thrown 1 sec later and both strikesimultaneously.This means that the time of travel for the second stone = 3 - 1 = 2 sec
Hence, 44.1 = 2u+129.8(2)2
⇒44.1−19.6=2u⇒u=12.25m/s
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:
A
Time 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=3sec
Second stone is thrown 1 sec later and both strikesimultaneously.This means that the time of travel for the second stone = 3 - 1 = 2 sec
Hence, 44.1 = 2u+129.8(2)2
⇒44.1−19.6=2u⇒u=12.25m/s
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