A ball is thrown from the top of a tower in vertically upward direction. The velocity at a point h meter below the point of projection is twice of the velocity at a point h meter above the point of projection. Find the maximum height reached by the ball above the top of tower.

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(4/3)h

(5/3)h

answer is C.

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Detailed Solution

$\begin{array}{l} H = \frac{u^{2}}{2 g}; given v_{2} = 2 v_{1} . . . . . . (i) \\ A to B : v_{1}^{2} = u^{2} - 2 g h . . . . . . . (i i) \\ A to C : v_{2}^{2} = u^{2} - 2 g (- h) . . . . . . . (i i i) \end{array}$

Solving (i), (ii) and (iii), we get the value of u² as 10gh/3 and then we get the value of H by using

$H = \frac{u^{2}}{2 g} (see figure) or H = \frac{5 h}{3}$

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