A particle falling vertically from a height hits a plane surface inclined to horizontal at an ange ` theta` with speed ` v_0` and rebounds elastically (Fig. 2 (RP). 28). Find the distance aling the plane where it will hit second time.
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Explanation

Considering x and y-axes as shown in the diagram.
For the motion of the projectile from O to A.
`y=0, u_(y) = v_(0) cos theta`
`a_(y)=-g cos theta, t=T`
Applying equation of kinematics,
`y=u_(y)t+(1)/(2)a_(y)t^(2)`
`rArr " " 0=v_(0)cos theta T + (1)/(2)(-g cos theta) T ^(2)`
`rArr " " T, [v_(0)cos theta -(g cos theta T)/(2)]=0`
`T=(2v_(0)cos theta)/(g cos theta)`
As T = 0, corresponds to point O
Hence, `" " T=(2v_(0))/(g)`
Now considering motion along OX.
`x=L, u_(x) =v_(0) sin theta, a_(x) = g sin theta, t = T= (2v_(0))/(g)`
Applying equati of kinematics,
`x=u_(x)t+(1)/(2) a _(x) t^(2)`
`rArr " " L=v_(0) sin theta t + (1)/(2) g sin theta t ^(2) = (v_(0) sin theta) (T) + (1)/(2) g sin theta T^(2)`
`=(v_(0) sin theta)((2v_(0))/(g))+(1)/(2) g sin theta xx((2v_(0))/(g))^(2)`
`=(2v_(0)^(2))/(g) sin theta +(1)/(2) g sin theta xx (4v_(0)^(2))/(g^(2))=(2v_(0)^(2))/(g)[sin theta + sin theta]`
`rArr " " L=(4v_(0)^(2))/(g)sin theta`