# Falling Objects| Kinematics| Boulder From Top of a Cliff| College Physics| Problem 2.53

### 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

We are given the following values:

$v_0=0\:m/s,\:y_0=0\:m,\:y=-250\:m,\:a=-9.80\:m/s^2$

We solve for the velocity at the bottom of the cliff using the equation

$v^2=\left(v_0\right)^2+2a\Delta y$

or we can rewrite it as

$v=\sqrt{\left(v_0\right)^2+2a\Delta y}$

Substitute the given values

$v=\sqrt{\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-250\:m\right)}=70.0\:m/s$

The boulder is traveling at about 70 m/s downward when it strikes the ground.

PART B

Let $t_s$ be the time for the sound to travel to the tourist

$t_s=\frac{height\:of\:cliff}{v_s}=\frac{250\:m}{335\:m/s}=0.7463\:s$

Let $t$ be the total time before the tourist can react

$t=t_s+reaction\:time=0.7643\:s+0.300\:s=1.046\:s$

Let $t_b$ be the time it takes the rock to reach the bottom

$t_b=\frac{v-v_o}{g}=\frac{-7.00\:m/s-0\:m/s}{-9.8\:m/s^2}=7.143\:s$

Now, subtract the times

$t_b-t=7.143\:s-1.046\:s=6.10\:s$

The tourist has to get out of the way in 6.10 seconds.