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When a ball is thrown up vertically with velocity $υ_{0}$ , it reaches a maximum height of $h$ . If one wishes to triple the maximum height then the ball should be thrown with velocity

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$3 υ_{0}$

$\frac{3}{2} υ_{0}$

$\sqrt{3} υ_{0}$

$9 υ_{0}$

answer is A.

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Detailed Solution

we know maximum height is given by

$h = \frac{u^{2}}{2 g}$

$u ∝ \sqrt{h}$

Therefore initial velocity should be $\sqrt{3} u_{0}$

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