### 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

Substitute the given values into the equation

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

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

Substitute the given values

Solve for t

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

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