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Question

# A stone is dropped by a person from the top of a tower, which is 200 m tall. At the same time another stone is thrown upwards, with a velocity of 50 msâˆ’1 by a person standing at the foot of the tower. Find the time after which two stones meet.

A
4 s
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B
5 s
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C
8 s
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D
10 s
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Solution

## The correct option is A 4 s Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.For the stone that is dropped its initial velocity u=0ms−1 ; displacement s = x and acceleration = acceleration due to gravity (g).Using s=ut+12at2,wegetx=(0)t+12gt2→(1).For the stone that is projected vertically upwards, its initial velocity, u=−50ms−1; displacement s =-(200-x) and acceleration a =g.using s=ut+12at2,weget−(200−x)=−50×t+1/2gt2200=50t−1/2gt2+x→(1)(2)Fromtheequations(1)and(2)wehave200=50t−1/2gt2+1/2gt2⇒200=50∴t=4s

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