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Question

Aball is droped from the top of the tower of height h .It covers a distance of h/2 in the last second of its motion .How long does the ball remain in air .(Take g = 10 ms2)

A
2 s
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B
(2+2) s
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C
2s
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D
None of the above
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Solution

The correct option is C (2+2) s
Let time during which ball remain in the air t,
From the kinematics,
s=ut+12gt2 (u=0m/s)
h=gt22. . . . .(1)
The distance travelled by the ball in nthsec is given by
Sn=u+g2(2n1)
at n=(t1)sec
h2=g2[2(t1)1]
h=g(2t3). . . . . (2)
By solving equation (1) and (2), we get
gt22=g(2t3)
t24t+6=0 (Quadratic equation)
t=221
By rationalisig,
t=221×2+12+1
t=(2+2)s
The correct option is B.

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