A ball is thrown horizontally at a speed `v_(0)` from the top of a hill whose slope is a (to the horizontal). Assuming the ball's impact on the hill's surface to be elastic find the point where it will hit the hill the second time.

Answer

The ratio of the distances is `t_(2)//t_(1) =1+ 2 sin^(2) alpha`.

Explanation

Arrange the coordinate axes as shown in Fig. 10.11. Then com ponants of the initial velocity along the axes will be `v_(0)x = v_(0) cos alpha , v_(0)y = v_(0) sin alpha` and the acceleration components will be `a_(x) = g sin alpha, a_(y) = -g cos alpha`. The equation of motion for the first part of the tra- jectory should be written thus:
`v_(x)=v_(0)xa_(x)t=v_(0)cosalpha+gtsinalpha`
`v_(y)=v_(0)y+a_(y)t=v_(0)sinalpha-gtcosalpha`
`x=x_(0)+v_(0)xt+(a_(x)t^(2))/2=v_(0)tcosalpha+(gt^(2)sinalpha)/2`
`y=y_(0)+v_(0)yt+(a_(y)t^(2))/2=v_(0)tsinalpha-(gt^(2)cosalpha)/2`
Since at the point `A_(1)` the cordinate `y_(1) = 0`, it follows that
`t_(1)=(2v_(0)tanalpha)/g,x_(1)=(2v_(0)^(2)sinalpha)/(gcos^(2)alpha)`
`v_(1x)=(v_(0))/(cosalpha)(1+sin^(2)alpha),v_(1y)-v_(0)sinalpha`
The longitudinal velocity component does not change- after an elastic impact, but the lateral component changes sign. Therefore

for the second part of the trajectory
`v_(0)x=v_(1)x=(v_(0))/(cosalpha)(1+sin^(2)alpha)(1+sin^(2)alpha),v_(0y)=-v_(1y)=v_(0)sinalpha`
The equation of motion, by analogy with the first case, will be
`x=x_1+v_(0x)t+(a_(x)t^(2))/2=(2v_(0)^(2)sinalpha)/(gcos^(2)alpha)+(v_(0)t(1+sin^(2)alpha))/(cosalpha)+(gt^(2)sinalpha)/2`
`y=y_(1)+v_(0y)^(2)+(a_(y)t^(2))/2=v_(0)tsinalpha-(gt^(2)cosalpha)/2`
At the point `A_(2)` the ordinate is again zero, i.e. `y_(2) = 0`, therefore the x-coordinate of the point `A_(2)` is
`x_(2)=(4v_(0)^(2)sin^(2)alpha)/(gcos^(2)alpha)(1+sin^(2)alpha)`
The ratio of displacements is
`(l_(2))/(l_(1))=(x_(2)-x_(1))/(x_(1))=1+2sin^(2)alpha`