Question

A projectile is fired with a speed $u$ at an angle $\theta$ above the horizontal floor. The coefficient of restitution between the projectile and the floor is $e$. Find the position from the starting point, where the projectile will land at its second collision.

• A. $\frac{e^2{u}^{2}sin2\theta }{g}$
• B. $\frac{(1-e^2)u^{2}sin2\theta }{g}$
• C. $\frac{(1-e)u^{2}sin\theta cos\theta }{g}$
• D. $\frac{(1+e)u^{2}sin2\theta }{g}$

Answer 1

The correct option is D $\frac{(1+e)u^{2}sin2\theta }{g}$

Range of projectile before first collision is $R=\frac{u^2\sin 2\theta }{g}$
After first collision, horizontal velocity remains same i.e $u\cos \theta$
Vertical velocity after first collision $=eu\sin \theta$

Now, time of flight after first collision is $T^{\prime }=\frac{2eu\sin \theta }{g}$
Range after first collision is
$R^{\prime }=\left(u\cos \theta \right)T^{\prime }=\frac{e{u}^2\sin 2\theta }{g}$
Hence,
$x=R+R^{\prime }=\frac{u^2\sin 2\theta }{g}(1+e)$