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Question

A stone dropped from topof 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(t2−t21)12g(t2−t22) or −t1t2=t2−t21t2−t22

By solving t =√t1t2


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