There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff.
(a) How fast will it be going when it strikes the ground?
(b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

Solution:

Part A

Consider the Figure A. We shall consider two position points — position 1 at the top of the cliff, and position 2 on the ground.

Position 1 is 250 meters above the ground (y=250 m), and since this is the initial position, t=0, and the initial velocity is v_y=0.

Position 2 is on the ground (y=0), and we do not know the time and velocity at this point.

For this part, we will solve for the value of v_y at position 2.

\begin{align*}
\left( v_{y_{2}} \right)^2 &  = \left( v_{y_{1}} \right)^2+2a \Delta y \\
\left( v_{y_{2}} \right)^2 &  = \left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m}-250\ \text{m} \right) \\
\left( v_{y_{2}} \right)^2 &  = 4905\ \text{m}^2/\text{s}^2 \\
v_{y_{2}} & = \pm \sqrt{4905\ \text{m}^2/\text{s}^2} \\
v_{y_{2}} & = \pm \ 70.0\ \text{m/s}\\
v_{y_{2}} & = -  70.0\ \text{m/s} \qquad {\color{gold}\left( \text{Answer} \right) }\\
\end{align*}

Since the boulder is moving downward at this point, we shall consider the negative sign of the velocity as an indication of the downward direction of the velocity.

To answer the question, the boulder is going at about 70.0 m/s downward when it strikes the ground.

Part B

Let:
t_s be the time for the sound to travel from the top of the cliff to the tourist;
t be the total time elapsed before the tourist can react;
t_b be the time for the rock to travel from the top of the cliff to the ground
t_t be the amount of time the tourist has to get out of the way after hearing the sound of the rock breaking loose

Since we are given the speed of sound, we have

\begin{align*}

t_s & = \frac{\text{height of the cliff}}{\text{speed of sound}} \\
t_s & = \frac{250\ \text{m}}{335\ \text{m/s}}\\
t_s & =0.746 \ \text{s}

\end{align*}

So, the tourist can react after

\begin{align*}

t & =t_s + \text{reaction time} \\
t & = 0.746\ \text{s}+0.300\ \text{s} \\
t &= 1.046\ \text{s}

\end{align*}

Then, we can compute for the total time it takes for the rock to travel from top to the ground.

\begin{align*}

t_b & =\frac{v_{y_2}-v_{y_1}}{a} \\
t_b & =\frac{-70\ \text{m/s}-0\ \text{m/s}}{-9.81\ \text{m/s}^2} \\
t_b & = 7.14\ \text{s}

\end{align*}

Therefore, the tourist still has the time to get out of the way. That is,

\begin{align*}

t_t & =7.14\ \text{s}-1.046\ \text{s} \\
t_t & = 6.09\ \text{s} \qquad {\color{gold}\left( \text{Answer} \right) } \\

\end{align*}

\therefore the tourist has about 6.09 seconds to get out of the way.