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From the top of a tower a stone is throw...

From the top of a tower a stone is thrown up which reaches the ground in time `t_(1)`. A second stone thrown down with the same speed reaches the ground in a time `t_(2)`. A third stone released from rest from the same location reaches the ground in a time `t_(3)`. then:

A

`(1)/(t_(3)) = (1)/(t_(1)) + (1)/(t_(2))`

B

`t_(3)^(2) = t_(1)^(2) - t_(2)^(2)`

C

`t_(3) = (t_(1) + t_(2))/(2)`

D

`t_(3) = sqrt(t_(1) t_(2))`

Text Solution

Verified by Experts

The correct Answer is:
D

`-H = ut_(1) - (1)/(2) g t_(1)^(2)`
`H = u t_(2) + (1)/(2) g t_(2)^(2)`
`H = (1)/(2) g t_(3)^(2)`
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