Correct Answer - A::B
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)`