From a cliff of height H = 5 m , a stone is thrown…

From a cliff of height $H = 5\ \mathrm{m}$, a stone is thrown horizontally with $v_0 = 25\ \mathrm{m/s}$. When it hits the water it bounces off at the same angle at which it arrived (without losing speed). Where does it hit the water the second time?

Explanation of Method

A stone skipping off the water.

Step 1: First impact

\[ t_1 = \sqrt{\frac{2H}{g}} = \sqrt{\frac{10}{9.81}} \approx 1.010\ \mathrm{s}, \]

\[ x_1 = v_0 t_1 \approx 25.2\ \mathrm{m}. \]

Its velocity then: $v_x = 25\ \mathrm{m/s}$, $v_y = \sqrt{2gH} \approx 9.90\ \mathrm{m/s}$ downward, at $\arctan\tfrac{9.90}{25} \approx 21.6^\circ$ to the water.

Step 2: The bounce

Reflection at the same angle means the vertical component reverses and the horizontal one stays: the stone leaves with $v_x = 25$, $v_y = +9.90\ \mathrm{m/s}$. The second arc is an ordinary symmetric throw:

\[ \Delta x = v_x \cdot \frac{2v_y}{g} = 25 \cdot \frac{2 \cdot 9.90}{9.81} \approx 50.5\ \mathrm{m}. \]

Step 3: Second impact

\[ x_2 = x_1 + \Delta x = 25.2 + 50.5. \]

\[x_2 \approx 75.7\ \mathrm{m} \ \text{from the foot of the cliff}\]

$5\ \mathrm{m}$ $25.2\ \mathrm{m}$ $75.7\ \mathrm{m}$ $21.6^\circ$

Vertical scale stretched 3×. The stone leaves the water at the same angle and speed at which it arrived, so the second arc is a symmetric throw that reaches the cliff's height again.

Result

\[\boxed{\,x_2 \approx 75.7\ \mathrm{m} \ \text{from the foot of the cliff}\,}\]

Remark

Note. The second arc is exactly twice as long as the first, because it is the first half-parabola plus its mirror image: $x_2 = 3\,v_0\sqrt{2H/g}$. The original answer, "about 77 m", came from rough rounding; the formula it used gives 75.7 m.