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Question

A ball is dropped from the top of a tower of height 80 m. At the same time, another ball is projected horizontally from the tower. Find the time taken by both the balls to reach the ground. (Take g = 10 $m s^{- 2}$ ).

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Solution

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Vertical velocity of both balls is initially is zero.
So, $s = u t + \frac{1}{2} a t^{2}$
$80 = s = 0 + \frac{1}{2} \times 10 \times t^{2}$
$16 = t^{2}$
t = 4 sec for both balls.

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