Ball Bouncing on Slope

If the slope is angled at $\theta$ to the horizontal, the ball experiences an acceleration of $g\cos\theta$ normal to the slope, and an acceleration of $g\sin\theta$ parallel to the slope. Thus the maximum distance $d$ of the ball from the slope is related to the component $v_\perp$ of the ball which is normal to the slope at the point of impact by the formula $v_\perp^2 \; = \; 2gd\cos\theta$ Since collisions with the slope are perfectly elastic, $v_\perp$ remains unchanged from one bounce to the next, and hence $d$ stays constant. Of course, the distance along the slope between impact points increases with time.

Consider the system in a frame of reference where the $x$ axis is parallel to the slope. In this frame of reference there is a "vertical" ( $y$ direction) acceleration of gravity (that is reduced by a factor of $\cos \alpha$ , where $\alpha$ is the angle of the slope relative to the horizontal direction). The equations of motions in the $x$ and $y$ directions are linear and independent. Accordingly, the ball will bounce to the same height in the $y$ direction after each collision, just it would be if dropped onto a horizontal surface. Additionally, the ball will be also subject to a force in the $x$ direction ( $F=mg \sin \alpha$ ), that will make it accelerate, so that the distance between the subsequent touching points on the slope will be larger and larger with time.

considering the two extremes, either the slope is vertical or horizontal, both give a constant h.

The question has a problem: the ball does not roll down the hill. “Continues down the hill” is probably better.

Conservation of energy is what came to mind for me first, and i guess it worked .