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Question

A ball is dropped from the top of a tower of height h. It covers a distance of h2 in the last second of its motion. How long does the ball remain in air?

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Solution

For 1sth2 fall
h2=0t1+12gt21[ from S=μt+12gt2]
h2=12gt21hg=t21....(1)
Also we know
V=0+gt1V=gt1
For 2ndh2 fall
h2=(gt1)t2+12gt2212.hg=t1t2+t222t212=t1t2+t222[from(1)]
Given that t2=1
t21=2t1+1t212t11=0t1=2±4+42=2±222t1=1+2
total time for which ball remains in air.
t1+t2=1+2+1=2+2seconds


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