A body falls from a height h onto an inclined plane with…

A body falls from a height $h$ onto an inclined plane with angle of inclination $\alpha$. Each time it hits the plane it bounces off elastically (the speed keeps its magnitude and the angle of reflection equals the angle of incidence). Find the distances $x_1, x_2, \dots, x_n$ between successive points where it hits the plane.

Explanation of Method

A body bouncing down an inclined plane.

Step 1: Speed at the first impact

\[ v_0 = \sqrt{2gh}, \ \text{directed vertically downward.} \]

Step 2: Axes along and across the plane

Take $x$ down the slope and $y$ perpendicular to the plane (outward). Gravity has components along both axes: $g_x = g\sin\alpha \ (\text{down the slope})$.

\[ g_y = -g\cos\alpha \ (\text{toward the plane}). \]

The falling body arrives with components $v_0\sin\alpha$ along the slope and $v_0\cos\alpha$ into the plane. The elastic bounce reverses only the perpendicular part: $v_x = v_0\sin\alpha$.

\[ v_y = +v_0\cos\alpha. \]

Step 3: Every hop lasts the same time

Across the plane, each hop starts with $v_0\cos\alpha$ and ends with it reversed, under the constant deceleration $g\cos\alpha$: $\tau = \frac{2v_0\cos\alpha}{g\cos\alpha} = \frac{2v_0}{g}$.

This is the same for every hop, so the bounces happen at equal time intervals.

Step 4: Motion along the slope

Along the slope the bounces change nothing: the body simply accelerates uniformly with $g\sin\alpha$, starting at $v_0\sin\alpha$. The $n$-th hop starts after $(n-1)\tau$, with speed

\[ u_n = v_0\sin\alpha + g\sin\alpha\,(n-1)\tau = (2n - 1)\,v_0\sin\alpha. \]

So the length of the $n$-th hop is

\[ x_n = u_n\tau + \tfrac12 g\sin\alpha\,\tau^2, \]

\[ x_n = (2n-1)\,v_0\sin\alpha\,\frac{2v_0}{g} + \tfrac12 g\sin\alpha\,\frac{4v_0^2}{g^2} \]

\[ x_n = \frac{4n\,v_0^2\sin\alpha}{g}. \]

With $v_0^2 = 2gh$: $x_n = 8\,n\,h\sin\alpha$.

So $x_1 = 8h\sin\alpha$, $x_2 = 16h\sin\alpha$, $x_3 = 24h\sin\alpha$, and so on.

Check

Second hop directly: $ x_2 = 3v_0\sin\alpha\,\tfrac{2v_0}{g} + \tfrac12 g\sin\alpha \left(\tfrac{2v_0}{g}\right)^2 $

$ x_2 = \tfrac{8v_0^2\sin\alpha}{g} = 16h\sin\alpha = 2 \cdot 8h\sin\alpha\ \checkmark $

Level ground ($\alpha = 0$): the body bounces straight up and down at one spot, $x_n = 0\ \checkmark$

$x_1 = 8\,h\sin\alpha$ $x_2 = 16\,h\sin\alpha$ $x_3 = 24\,h\sin\alpha$ $h$ $\alpha$

Drawn to scale for $\alpha = 20^\circ$. Every hop lasts the same time $2v_0/g$ and rises to the same height above the plane, but gravity keeps speeding the body up along the slope, so the hops get longer in the ratio 1 : 2 : 3.

Result

\[\boxed{\,x_n = 8\,n\,h\sin\alpha\,}\]

Remark

About the original solution. Its result $x_n = 8h\sin^2\alpha$ is wrong: it used $v_0\sin\alpha$ for the time of flight and ignored that the along-slope speed grows from hop to hop, so every hop came out the same. The hint expression given in the problem is correct (apart from a missing $\tfrac12$ in front of the $g\sin\alpha$ term and the typo $v_6$ for $v_0$): it gives $x_2 = 2 \cdot 8h\sin\alpha$, which agrees with $x_n = 8nh\sin\alpha$. The distances grow in the ratio 1 : 2 : 3 : …, and each hop is $8h\sin\alpha$ longer than the previous one.