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(C) The maximum height reached by a ball is $h = 5 \, m$.The velocity of projection $u$ is given by $u = \sqrt{2gh}$.Substituting the values: $u = \sq

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$60$

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(C) The maximum height reached by a ball is $h = 5 \, m$.The velocity of projection $u$ is given by $u = \sqrt{2gh}$.Substituting the values: $u = \sqrt{2 	imes 10 	imes 5} = \sqrt{100} = 10 \, m/s$.The time taken to reach the maximum height (time of ascent) is $t = \frac{u}{g}$.Substituting the values: $t = \frac{10}{10} = 1 \, s$.Since the next ball is thrown when the previous one is at its maximum height,the time interval between two consecutive throws is $1 \, s$.Number of balls thrown per minute = $\frac{60 \, s}{1 \, s/	ext{ball}} = 60 \, 	ext{balls/min}$.

$A$ very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5 \, m$,the number of balls thrown per minute is (take $g = 10 \, m/s^2$).

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