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?

Answer

A, B

Explanation

The projected velocity =u
The angle of prjection `=theta`
Whentehprojectilehits the ground for the 1st time,
The velocity would be the same i.e u
Here the component of velocity paralel to ground
`u costheta` should remain constant
But teh verticla component of the projectile under goes.

`rarr v=eusintheta`
Now for the 2nd projectile motion.
`u=velocity of projection
`=sqrt((ucostheta)^2+(eusintheta)^2)`
and Angle of projection
`=alpha=tan^I-1((eusintheta)/(ucostheta))`
`=tan^-1 (etantheta)`
`or taN/Alpha=etantheta`..............2
Because `y=xtaN/Alphatan^-1(gx^2sec^2alpha)/(2u^2)`............e
`Here y=0, taN/Alpha`
`=etantheta`
`sec^2alha=1+e^2tan^2theta`
and `u^2=u^2cos^2theta+e^2u^2sin^2theta`
putting the above values in the equation 3
`xetantheta=(gxx^2(1+e^2tan^2theta))/(2u^2(cos^2theta+e^2sintheta))`
or x=(2eu^2tantheta(cos^2theta+e^2sin^2theta))/(g(1+e^2tan^2theta))`
`=(2eu^2tantheta.cos^2theta)/g`
`(eu^2sin2theta)/g`
`rarr So, from the starting point O it will fasll at a distance
`(u^2sin2theta)/g+(eu^2.sin2theta)/g`
`(u^2sin2theta)/g(1+e)`