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(B) Let the initial velocity be $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$. Substituting $h = 25 \ m$ and $g = 10 \ m \ s^{-2}$,we get

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$30$

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(B) Let the initial velocity be $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$. Substituting $h = 25 \ m$ and $g = 10 \ m \ s^{-2}$,we get $25 = ut - 5t^2$,which simplifies to $5t^2 - ut + 25 = 0$. Let $t_1$ and $t_2$ be the two times at which the ball is at height $25 \ m$. The sum of roots $t_1 + t_2 = \frac{u}{5}$ and the product of roots $t_1 t_2 = \frac{25}{5} = 5$. The time interval is $|t_2 - t_1| = 4 \ s$. Using the identity $(t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1 t_2$,we have $4^2 = (\frac{u}{5})^2 - 4(5)$. $16 = \frac{u^2}{25} - 20$,so $\frac{u^2}{25} = 36$. $u^2 = 36 	imes 25 = 900$,which gives $u = 30 \ m \ s^{-1}$.

If a ball projected vertically upwards with a certain initial velocity from the ground crosses a point at a height of $25 \ m$ twice in a time interval of $4 \ s$,then the initial velocity of the ball is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

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