Question

20. A ball is thrown horizontally from the top of a vertical tower of height H with a velocity of 2gH .
Magnitude of velocity of the ball when it strikes the ground is

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Answer 1

Explanation:

HEY MATE ..........

$given \\ \\ horizontal \: velocity = \sqrt{2gh} \\ \\ vertical \: velocity = 0 \\ \\ using \: \\ \\ s = ut + \frac{1}{2} a {t}^{2} \\ \\ = - h = 0 - \frac{1}{2} g {t}^{2} \\ \\ = t = \sqrt{ \frac{2h}{g} } \\ \\ = x = horizontal \: velocity \: \times t \\ \\ = x = \sqrt{2gh } \times \sqrt{ \frac{2h}{g} } = 2h$

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