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Question

A body is thrown upward at an angle θ with horizontal from a point A on top of a tower. It reaches ground in time t1. If it is thrown downward at an angle θ with horizontal from same point with same speed, it reaches the ground in time
t2. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

A
(t1+t2]2
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B
(t1+t2]2 sinθ
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C
t1t2
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D
2t1t2t1+t2
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Solution

The correct option is C t1t2
If velocity of the body is u0, then vertical component of velocity will be u0 sinθ .
Sign convention - Taking upward direction from point A as positive .
When ball is thrown upward h = (u0sinθ)t1 gt122 (i)
When ball is thrown downward h = (u0sinθ)t2 gt222 (ii)
Subtracting (ii) from (i) , we get 0 = (u0sinθ)(t1 + t2) + g(t22t12)2 u0sinθ = g(t1t2)2.
Putting above in equation (ii), we get h = g(t1t2)t22 gt222 h=gt1t22 .
For free fall initial velocity = 0 h = gt22 .
Equating h, we get gt1t22 = gt22 t = t1t2

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