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