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A ball is thrown up at an angle with the horizontal. Then the total change of momentum by the instant it returns to ground is:

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Weight of the ball $\times$ half the time of flight

Acceleration due to gravity $\times$ total time of flight

Weight of the ball $\times$ total time of flight

Weight of the ball $\times$ horizontal range

answer is C.

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Detailed Solution

At the instant of projection $\vec{P_{x}} = m \vec{u_{x}}$ , $\vec{P_{y}} = m \vec{u_{y}}$

At the instant of returning to the ground

$\vec{P_{x}} = m \vec{u_{x}}$ , $\vec{P_{y}} = - m \vec{u_{y}}$

Change in momentum

$| \vec{Δ p} | = 2 m u_{y}$

$= 2 m u \sin θ$

$= m g \times \frac{2 u \sin θ}{g}$

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