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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 ms1 by a person standing at the foot of the tower. Find the time after which two stones meet.
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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=0ms1 ; 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=50ms1; displacement s =-(200-x) and acceleration a =g.
using s=ut+12at2,weget(200x)=50×t+1/2gt2200=50t1/2gt2+x(1)(2)Fromtheequations(1)and(2)wehave200=50t1/2gt2+1/2gt2200=50t=4s

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