# College Physics 2.53 – A boulder breaking loose from the top of a cliff

## 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.