Question

A ball falls freely a height onto and smooth inclined plane forming an angle $\alpha$ with the horizontal. If the ratio of the distance between the points at which the jumping ball strikes the inclined plane is 1:2:x, find the value of x. Assume the impacts to be elastic.

Answer 1

Let $x_1{x}_2{x}_3$ be the point of collision of three consecutive jump ball approaches with velocity v from point B in every jump time taken is same say (+).

$x_1=\left(v\cos \alpha \right)t+\frac{1}{2}g\sin \alpha t^2\to \left(1\right)$

at point B velocity in x-direction is

$v^{{}^{\prime }}=(v\cos \alpha +g\sin \alpha t)$

$x_2=v^{\prime }t+\frac{1}{2}g\sin \alpha t^2=(v\cos \alpha +g\sin \alpha t)t+\frac{1}{2}g\sin \alpha t^2\to \left(ii\right)$

given that $\frac{x}{x_2}=\frac{1}{2}$

Therefore $g\sin t^2=v\cos \alpha t+\frac{1}{2}g\sin \alpha t^2$

at point of c velocity in x-direction will be

$g\sin t^2=v\cos \alpha t+\frac{1}{2}g\sin \alpha t^2{v}^{\prime \prime }=(v\cos \alpha +2g\sin \alpha t)x_3=v"t+\frac{1}{2}g\sin \alpha t^2=(v\cos \alpha +2g\sin \alpha t)t+\frac{1}{2}g\sin \alpha t^2{x}_3=3\left(v\cos \alpha +\frac{1}{2}g\sin \alpha t^2\right)\to \left(iii\right)$

from equation (1) and (3)

$\frac{x_1}{x_3}=\frac{1}{3}$

therefore ratio $x:x_2:x_3=1:2:3$