A ball is dropped from a height. If it taken 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take $g = 10 m / s^{2} ?$

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answer is 48.

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Detailed Solution

$if height of the tower is h and total time taken is t$

$h = \frac{1}{2} {gt}^{2} \to \to (1)$

$h - 6 = \frac{1}{2} g {(t - 0.2)}^{2} \to \to \to (2)$

$Solving (1) and (2) we get t = 3.1$

$substituting the value of t in equation (1)$

$h = 48.01 m$

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