Question

A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the playerhears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air is 343 m/s.

Short answer

Initial speed given to the rock is 9.29 m/s.

Step-by-step solution

Time taken for the rock to reach water surface

Taking vertical components only

$\begin{array}{rcl}y& =& y_i+u_{y}t+\frac{1}{2}{a}_y{t}^2\\ 0& =& 40+0-\frac{1}{2}(9.8)t^2\\ 4.9t^2& =& 40\\ t& =& 2.86\mathrm{s}\end{array}$

Since the total time taken is given as 3s.

Hence, the time in which the sound wave comes back to the player $\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{86}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{}\mathbf{s}$.

Using Pythagorean Theorem to calculate the horizontal distance

Since, the sound wave travels back in a straight line, we can use Pythagorean Theorem for x,

$$\begin{gathered} x=\sqrt{{(48.02)}^2-{\left(40\right)}^2} \\ x=26.57\mathrm{m} \end{gathered}$$

Solving for the initial speed

We know,

$$\begin{gathered} x=u_{x}t \\ {u}_x=\frac{26.57}{2.86} \\ {u}_x=9.29\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, Initial speed given to the rock is$9.29\mathrm{m}/\mathrm{s}$.