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Q.

A ball is thrown vertically upward from the roof of a building with a certain velocity. It reaches the ground in 9 s. When it is thrown downward from the roof with thesame initial speed, it takes 4 s to come at ground. How much time (in second) will it take to reach at ground if it just released from the rest from the roof?

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answer is 6.

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Detailed Solution

Let the height of building is h. Event I, −h=v0t1−12gt12  ..........(i) Event II, h=v0t2+12gt22    ............(ii) Event III, h=12gt2   ...........(iii)On multiplying Eq. (i) by t2 and Eq. (ii) by t1 and add, we get ht1+t2=12gt1t2t1+t2 Or  12gt2=12gt1t2∴ t=t1×t2=9×4=6s
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