Final answer:
The rock is 93.975 m above the hiker when he can see it and he has approximately 3.13 seconds to move before the rock hits his head. Therefore, the correct answer is b) 98.1 m.
Step-by-step explanation:
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. The hiker 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?
To solve part (a), we need to determine how far the rock has fallen in the time before the hiker sees it. We can use the equation for the distance fallen due to gravity, which is d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.80 m/s²), and t is the time in seconds.
Plugging the values in gives us d = 1/2 * 9.80 m/s² * (1.50 s)^2, which calculates to d = 11.025 m. Therefore, the rock was 105 m - 11.025 m = 93.975 m above the hiker when he can see it.
(b) How much time does he have to move before the rock hits his head?
To solve part (b), we need to find the total time it takes for the rock to fall 105 m. Starting from rest, we can use the equation t = √(2 * h / g), where h is the height and g is acceleration due to gravity. With h = 105 m and g = 9.80 m/s², we get t = √(2 * 105 m / 9.80 m/s²) which is approximately 4.63 seconds. The hiker first sees the rock after 1.50 s, so he has 4.63 s - 1.50 s = 3.13 seconds to move out of the way.