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A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity, the impulse received by the ball due to gravity force during its flight is

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$4 \sqrt{m^{2} gh}$

$\sqrt{4 m^{2} gh}$

$\sqrt{8 m^{2} gh}$

$\sqrt{2 m^{2} gh}$

answer is C.

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Detailed Solution

As the ball reaches a height 'h',

The velocity of the ball at the ground level will be $\sqrt{2 gh}$

The change in velocity = 2mu

Impulse $= 2 mu = 2 m \sqrt{2 gh} = \sqrt{8 m^{2} gh}$

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