A) acceleration due to gravity x total time of flight
B) weight of the ball x half the time of flight
C) weight of the ball x total time of flight
D) weight of the ball x horizontal range
Correct Answer: C
Solution :
Change in momentum of the ball \[=mv\sin \theta -(-mv\sin \theta )=2mv\sin \theta \] \[=mg\times \frac{2v\sin \theta }{g}\] = weight of the ball \[\times \] total time of flight
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