Answer to Question #264816 in Physics for Bob626

Question

A soccer player kicks a rock horizontally off a 31.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.4 seconds later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Answer 1

The distance $L$ travelled by the sound is given as follows (from the right triangle):

$$L^2 = h^2 + d^2$$

The time of fall can be found from the kinematic equation:

$$h = \dfrac{gt_f^2}{2}\\ t_f = \sqrt{\dfrac{2h}{g}}$$

The time of sound propagation is then $t-t_f$ , where $t = 3.4s$, and $L$ can also be given as follows:

$$L = c(t-t_f)$$

where $c = 343m/s$ is the speed of sound.

During the time of fall, the rock travelled the following horizontal distance:

$$d = v_0t_f$$

where $v_0$ is its initial speed.

Substituting all these expressions into the first equation and expressing $v_0$, obtain:

$$c^2(t-t_f)^2 = h^2 + v_0^2t_f^2\\ v_0^2 = \dfrac{c^2(t-t_f)^2-h^2}{t_f^2}\\ v_0 = \sqrt{\dfrac{g\left[c^2\left( t-\sqrt{\frac{2h}{g}} \right)^2-h^2\right]}{2h}}=\\ =\sqrt{\dfrac{9.8\left[343^2\left( 3.4-\sqrt{\frac{2\cdot 31}{9.8}} \right)^2-31^2\right]}{2\cdot 31}}\approx 120m/s$$

Answer. 120m/s.