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(B) Let the first ball dropped from the top travel a distance $h_1$ in time $t$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$

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$2$

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(B) Let the first ball dropped from the top travel a distance $h_1$ in time $t$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get:$h_1 = \frac{1}{2}gt^2$Let the second ball thrown from the bottom travel a distance $h_2$ in the same time $t$. Using $s = ut - \frac{1}{2}gt^2$,where $u = 50 \, m/s$,we get:$h_2 = 50t - \frac{1}{2}gt^2$Since the total height of the tower is $100 \, m$,the sum of the distances covered by both balls when they meet is equal to the height of the tower:$h_1 + h_2 = 100$Substituting the expressions for $h_1$ and $h_2$:$\frac{1}{2}gt^2 + (50t - \frac{1}{2}gt^2) = 100$$50t = 100$$t = 2 \, s$Thus,the balls will cross each other after $2 \, s$.

$A$ ball is dropped from the top of a tower of $100 \, m$ height. Simultaneously,another ball is thrown upward from the bottom of the tower with a speed of $50 \, m/s$ $(g = 10 \, m/s^2)$. They will cross each other after . . . . . . $s$.

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