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Question

A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1 and t2 seconds respectively, then

A
t = t1 - t2
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B
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C
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D
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Solution

The correct option is C

If a stone is dropped from height h
then h=12gt2 ....(i)
If a stone is thrown upward with velocity u then,
h=ut1+12gt21 ....(ii)
If a stone is thrown downward with velocity u then
h=ut2+12gt22 ...(iii)
From (i) (ii) and (iii) we get
ut1+12gt21=12gt2 ..(iv)
ut2+12gt22=12gt2 ..(v)
Dividing (iv) and (v) we get
ut1ut2=12g(t2t21)12g(t2t22)
or t1t2=t2t21t2t22
By solving t=t1t2


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