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(D) Let the initial velocity of the ball be $u$.The time taken by a ball to reach its highest point is given by $t = \frac{u}{g}$.Since the juggler th

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$g / 2n^{2}$

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(D) Let the initial velocity of the ball be $u$.The time taken by a ball to reach its highest point is given by $t = \frac{u}{g}$.Since the juggler throws $n$ balls per second,the time interval between two consecutive throws is $T = \frac{1}{n}$.According to the problem,the next ball is thrown when the first ball reaches its highest point. Therefore,the time interval between throws must be equal to the time taken to reach the highest point:$T = t \implies \frac{1}{n} = \frac{u}{g}$.Solving for $u$,we get $u = \frac{g}{n}$.The maximum height $H_{\max}$ reached by the balls is given by the formula $H_{\max} = \frac{u^{2}}{2g}$.Substituting the value of $u$:$H_{\max} = \frac{(\frac{g}{n})^{2}}{2g} = \frac{g^{2} / n^{2}}{2g} = \frac{g}{2n^{2}}$.

$A$ juggler throws balls vertically upwards with the same initial velocity in the air. When the first ball reaches its highest position,he throws the next ball. Assuming the juggler throws $n$ balls per second,the maximum height the balls can reach is

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