While making some observations at the top of the 66 m tall Astronomy tower, Ron accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved