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A ball is thrown from the top of a tower in vertically upward direction. 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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(5/3)h

(4/3)h

answer is C.

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Detailed Solution

velocity at B if it is $v_{1}$ and at C if it is $v_{2}$ then $v_{2} = 2 v_{1} and distance travelled is BC = 2 h$

$according to kinematics equation$

${(v_{2})}^{2} - {(v_{1})}^{2} = 2 g (2 h) 3 v_{1}^{2} = 4 gh$

$maximum height above B is \frac{v_{1}^{2}}{2 g} = \frac{4 gh}{3 (2 g)} = \frac{2 h}{3}$

$t otal height above A is h + \frac{2 h}{3} = \frac{5 h}{3}$

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