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Q.

A juggler throws balls vertically up in the air. He throws one up whenever the previous one is at its highest point. If he throws n balls in each second, the height to which each ball rises above the point of throw will be (g is acceleration due to gravity)

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a

g2n2

b

2gn2

c

4gn2

d

g4n2

answer is A.

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Detailed Solution

The time interval is (1/n) s. In this time the ball rises/falls the required distance. Using s = ut + (1/2)at2, h = (1/2)g(1/n)2

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