# Falling Objects| Kinematics| Rock from a cliff| College Physics| Problem 2.51

### Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

PART A

The rock is from a 105-meter cliff. Since the rock break loose, the initial velocity is 0. After 1.5 seconds, we can solve for the distance traveled by the rock using the equation

$y=v_0t+\frac{1}{2}at^2$

Substitute the given values into the equation

$y=0+\frac{1}{2}\left(-9.80\:m/s\right)\left(1.5\:s\right)^2=-11.025\:m$

The negative sign of the distance tells us that the movement is downward. So, the rock has traveled 11.025 meters downwards. Therefore, the distance of the rock to the hiker is

$y=105\:m-11.025\:m=93.975\:m$

The rock is 93.975 meters from the hiker when he finally sees it.

PART B

So, we need to determine the time when the rock has traveled 105 meters. We shall use the formula

$y=v_0t+\frac{1}{2}at^2$

Substitute the given values

$-105\:m=0+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$

Solve for t

$t^2=\frac{-105\:m}{-4.90\:m/s^2}$

$t^2=\frac{150}{7}$

$t=\sqrt{\frac{150}{7}}=4.63\:s$

The rock will take 4.63 seconds to travel from the starting point to the ground. So, if the rock has already traveled for 1.5 seconds, the hiker has still the time to move out. The time is

$t_{move\:out}=4.63\:s-1.50\:s=3.13\:s$

The hiker has 3.13 seconds to move out to avoid getting hit by the rock.

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