A projectile is fired with a speed u at an angle $\theta$ above a horizontal field.The coefficient of restitution of collision between the projectile and the field is e.How far from the starting point,does the projectile makes its second collision with the field ?

The projected velocity=u

The angle of projection=$\theta$

When the projectile hits the ground for the 1st time,

The velocity would be the same i.e. u.

Here the component of velocity parallel to ground,

should remain constant.

But the vertical compoenent of the projectile undergoes.

$\Rightarrow v=eusin\theta$

Now for the 2nd projectile motion.

u=velocity of projection

$=\sqrt{(ucos\theta {)}^2+(eusin\theta {)}^2}$

and Angle of projection

$=\alpha =ta{n}^{-1}\left(\frac{eusin\theta }{ucos\theta }\right)$

$=ta{n}^{-1}(etan\theta$

or $tan\alpha =etan\theta$ ....(2)

Because, $y=xtan\alpha ta{n}^{-1}\frac{g{x}^{2}se{c}^2\alpha }{2u^2}$ ....(3)

Here, $y=0,tan\alpha$

$=etan\theta$

$se{c}^2\alpha =1+e^{2}ta{n}^2\theta$

and $u^2=u^{2}co{s}^2\theta +e^2{u}^{2}si{n}^2\theta$

Putting the above values in the equation (3),

$xetan\theta =\frac{g{x}^2(1+e^{2}ta{n}^2\theta )}{2u^2(co{s}^2\theta +e^{2}si{n}^2\theta )}$

$orx=\frac{2e{u}^{2}tan\theta \left(\right(co{s}^2\theta +e^{2}si{n}^2\theta )}{g(1+e^{2}ta{n}^2\theta )}$

$=\frac{2e{u}^{2}tan\theta .co{s}^2\theta }{g}$

$=\frac{e{u}^{2}sin2\theta }{g}$

$\Rightarrow$ So,from the starting point O,it will fall at a distance

$=\frac{u^{2}sin2\theta }{g}+\frac{e{u}^2.sin2\theta }{g}$

$=\frac{u^{2}sin2\theta }{g}(1+e)$