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Question

Change in momentum of the ball,

$\Delta \mathrm{p}=\mathrm{mv} \sin 	heta-(-\mathrm{mv} \sin 	heta)=2 \mathrm{mv} \sin 	heta$

Where $\mathrm{

Vedclass · Accepted Answer

weight of the ball  $	imes$ total time of flight

Vedclass · Answer

Change in momentum of the ball,

$\Delta \mathrm{p}=\mathrm{mv} \sin 	heta-(-\mathrm{mv} \sin 	heta)=2 \mathrm{mv} \sin 	heta$

Where $\mathrm{v}$ is the velocity of prjection and $	heta$ is the angle of projection with the horizontal and $\mathrm{m}$ is the mass of the ball.

$\Delta p=2 m g v 	imes \frac{\sin 	heta}{g}$

$=m g 	imes \frac{2 v \sin 	heta}{g}$

$=$ weight of the ball $	imes$ total time of flight.

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

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