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(C) The maximum height $h$ reached by a ball thrown with initial velocity $u$ is given by $h = \frac{u^2}{2g}$.Given $h = 5 \ m$ and taking $g = 10 \

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$60$

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(C) The maximum height $h$ reached by a ball thrown with initial velocity $u$ is given by $h = \frac{u^2}{2g}$.Given $h = 5 \ m$ and taking $g = 10 \ m/s^2$,we have $5 = \frac{u^2}{2 	imes 10}$,which gives $u^2 = 100$,so $u = 10 \ m/s$.The time taken to reach the maximum height (time of ascent) is $t = \frac{u}{g} = \frac{10}{10} = 1 \ s$.Since the next ball is thrown when the previous one is at its maximum height,the time interval between throwing two consecutive balls is $1 \ s$.Therefore,the number of balls thrown per minute is $\frac{60 \ s}{1 \ s/	ext{ball}} = 60 \ 	ext{balls/min}$.

Balls are thrown vertically upward 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 will be

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