Question

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?

How far above the hiker is the rock when he can see it?
1. d = vᵢt + frac{1}{2}gt²
2. d = vᵢt + gt²
3. d = vᵢt + frac{1}{2}gt²
4. d = vᵢt + gt

Answer 1

The rock is 93.975 m above the hiker when he sees it, and he has approximately 3.13 seconds to move out of the way before it hits the ground.

Step-by-step explanation:

We are given a scenario where a hiker at the base of a cliff in Mt. Arapiles witnesses a rock fall from a height of 105 m, but only sees it after a delay of 1.50 seconds. We need to solve for two things: (a) the height of the rock above the hiker when he first sees it, and (b) how much time he has to move to safety before it hits.

Part (a): Height of the rock when first seen by the hiker

We consider the acceleration due to gravity to be 9.8 m/s2. The formula to use is d = vit + (1/2)gt2 where vi is the initial velocity (0 m/s), t is the time, and g is the acceleration due to gravity.

Calculating we get:

d = 0 m/s * 1.5 s + (1/2) * 9.8 m/s2 * (1.5 s)2

= 0 + 0.5 * 9.8 * 2.25

= 11.025 m

The rock has fallen 11.025 m after 1.5 seconds, so the height above the hiker when he sees it is:

105 m - 11.025 m = 93.975 m

Part (b): Time left before the rock hits the hiker

Using the formula for the total time to hit the ground: t_total = √(2 * height / g)

We find:

t_total = √(2 * 105 m / 9.8 m/s2)

= √(210 / 9.8)

= √(21.42857)

= 4.63 s (approximately)

The rock takes 4.63 s to hit the ground, so the hiker has time left after seeing the rock:

Time left = 4.63 s - 1.50 s

= 3.13 seconds