A very large number of balls are thrown verti | vedclass

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(C) The time taken for a ball to reach the maximum height $h$ when thrown with initial velocity $u$ is given by $t = \frac{u}{g}$.At maximum height,th

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$60$

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(C) The time taken for a ball to reach the maximum height $h$ when thrown with initial velocity $u$ is given by $t = \frac{u}{g}$.At maximum height,the final velocity is $0$. Using $v^2 = u^2 - 2gh$,we get $0 = u^2 - 2gh$,so $u = \sqrt{2gh}$.Substituting $u$ into the time equation: $t = \frac{\sqrt{2gh}}{g} = \sqrt{\frac{2h}{g}}$.Given $h = 5\, m$ and $g = 10\, ms^{-2}$,the time taken is $t = \sqrt{\frac{2 	imes 5}{10}} = \sqrt{1} = 1\, s$.Since the next ball is thrown when the previous one reaches its maximum height,one ball is thrown every $1\, s$.Therefore,the number of balls thrown per minute $(60\, s)$ is $\frac{60\, s}{1\, s/	ext{ball}} = 60\, 	ext{balls/minute}$.

$A$ 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 $g = 10\, ms^{-2}$):-

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