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 $$ms^{-2}$$).

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Solution

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Vertical velocity of both balls is initially is zero. So, $$s = ut + \dfrac{1}{2}at^2$$ $$80 = s = 0 + \dfrac{1}{2}\times10\times t^2$$ $$16 = t^2$$ t = 4 sec for both balls.

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