MCQ
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
  • A
    acceleration due to gravity $\times$ total time of flight
  • B
    weight of the ball $\times$  half the time of flight
  • weight of the ball  $\times$ total time of flight
  • D
    weight of the ball  $\times$ horizontal range

Answer

Correct option: C.
weight of the ball  $\times$ total time of flight
c
Change in momentum of the ball,

$\Delta \mathrm{p}=\mathrm{mv} \sin \theta-(-\mathrm{mv} \sin \theta)=2 \mathrm{mv} \sin \theta$

Where $\mathrm{v}$ is the velocity of prjection and $\theta$ is the angle of projection with the horizontal and $\mathrm{m}$ is the mass of the ball.

$\Delta p=2 m g v \times \frac{\sin \theta}{g}$

$=m g \times \frac{2 v \sin \theta}{g}$

$=$ weight of the ball $\times$ total time of flight.

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