A particle of mass m is projected with a velocity v 0 at an angle α with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is e . The time after which the particle will stop bouncing off the ground is

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Aicharya · Accepted Answer

After each collision, the particle loses some part of its vertical velocity while its horizontal velocity remains unchanged.

$T = \frac{2 v_{0} s i n α}{g} + \frac{2 (e v_{0}) s i n α}{g} + \frac{2 (e^{2} v_{0}) \sin α}{g} + . . . . . . . . . .$

$\Rightarrow T = \frac{2 v_{0} s i n α}{g} (1 + e + e^{2} + . . . . . . . . .)$

$T = \frac{2 v_{0} s i n α}{g (1 - e)}$

A particle of mass m is projected with a velocity v 0 at an angle α with the horizontal. The coefficient of…

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