A ball is projected with initial velocity u at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is e, ______

e \frac{u^{2} sin 2 θ}{g}

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(1 + e) \frac{u^{2} sin 2 θ}{g}

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(1 + e + e^{2}) \frac{u^{2} sin 2 θ}{g}

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(1 - e) \frac{u^{2} sin 2 θ}{g}

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Solution

The correct option is C $(1 + e + e^{2}) \frac{u^{2} sin 2 θ}{g}$
Range of the projectile with velocity $u$ is given by $R = \frac{u^{2} sin 2 θ}{g}$

so before first collision $R_{1} = \frac{u^{2} sin 2 θ}{g}$

now vertical velocity changes to $e u_{y}$ as it is given that coefficient of restitution is e.

so before second collision $R_{2} = \frac{e u^{2} sin 2 θ}{g}$

similary before 3rd collision $R_{3} = \frac{e^{2} u^{2} sin 2 θ}{g}$

Total displacement $R = R_{1} + R_{2} + R_{3}$

$R = \frac{u^{2} sin 2 θ}{g}$ + $\frac{e u^{2} sin 2 θ}{g}$ + $\frac{e^{2} u^{2} sin 2 θ}{g}$

$R = (1 + e + e^{2}) \frac{u^{2} sin 2 θ}{g}$

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A ball is projected with initial velocity u at an angle θ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is e, ______

A ball is projected with initial velocity u at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is e, ______