A ball is thrown vertically upwards from the top of a tower. Velocity at a point ‘h’m vertically below the point of projection is twice the downward velocity at a point ‘h’m vertically above the point of projection. The maximum height reached by the ball above the top of the tower is

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$\frac{4}{3} h$

$\frac{5 h}{3}$

answer is D.

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Detailed Solution

v₁ = 2v₂
v₂ – u₂ = 2as
$\begin{array}{l} \sqrt{u^{2} + 2 gh} = 2 [\sqrt{u^{2} - 2 gh}] \\ \Rightarrow h = \frac{3}{5} \cdot \frac{u^{2}}{2 g} \Rightarrow h = \frac{3}{5} \cdot H \\ \Rightarrow H = \frac{5 h}{3} \end{array}$

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