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Question

An object thrown upwards with a velocity u from the top of a tower of height H reaches the ground in t1 seconds. When thrown downwards with the same speed it reaches the ground in t2 seconds. When dropped from the top of tower it reaches in t3 seconds. Then t3 is equal to (in terms of t1 and t2)

A
t1 + t2
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B
t1t2
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C
t1 - t2
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D
t1/t2
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Solution

The correct option is B t1t2
assuming downward direction as positive we get
H=ut1+12at21
and H=ut2+12at22
multiplying 1 with t2 and 2 with t1 and then add both we get
=>H(t1+t2)=12at1t2(t1+t2)
=>2H=aT1.t2
so when ball is simply released from top we get
H=12at23
using value of H we get
=>t23=t1.t2
so best answer is option B.

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