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(C) Let the two balls meet at point $P$ after time $t$.For ball $A$ dropped from the top,the distance covered downwards is $h_1 = \frac{1}{2}gt^2$.For

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$80$

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(C) Let the two balls meet at point $P$ after time $t$.For ball $A$ dropped from the top,the distance covered downwards is $h_1 = \frac{1}{2}gt^2$.For ball $B$ thrown upwards from the base,the distance covered upwards is $h_2 = ut - \frac{1}{2}gt^2$,where $u = 50 \ m/s$.The total height of the tower is $h_1 + h_2 = 400 \ m$.Substituting the expressions: $\frac{1}{2}gt^2 + (ut - \frac{1}{2}gt^2) = 400$.This simplifies to $ut = 400$.Given $u = 50 \ m/s$,we have $50t = 400$,which gives $t = 8 \ s$.The height from the base is the distance covered by ball $B$,which is $h_2 = ut - \frac{1}{2}gt^2$.Assuming $g = 10 \ m/s^2$,$h_2 = 50(8) - \frac{1}{2}(10)(8^2) = 400 - 5(64) = 400 - 320 = 80 \ m$.

$A$ man drops a ball from the roof of a tower of height $400 \ m$. At the same time,another ball is thrown upwards with a velocity of $50 \ m/s$ from the base of the tower. At what height from the base of the tower will they meet (in $m$)?

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