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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

$\frac{g}{2 n^{2}}$

b

$\frac{2 g}{n^{2}}$

c

$\frac{4 g}{n^{2}}$

d

$\frac{g}{4 n^{2}}$

answer is A.

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Detailed Solution

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

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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)