While making some observations at the top of the 63 m tall Astronomy tower, Ron accidently knocks a 0.5 kg stone over the edge. How fast will the stone be moving when
it hits the ground?

Analysing the Question:

We are given:

height of the tower (h) = 63 m

mass of stone (m) = 0.5 kg

initial vertical velocity of the stone (u) = 0 m/s  [ since the stone fell, there was no force applied on it to go downwards]

final vertical velocity of the stone (v) = v m/s

acceleration of the stone due to gravity (a) = 10 m/s²    [we are letting g = 10]

** We will be neglecting air resistance in this answer**

Finding the final velocity of the stone:

from the third equation of motion

v² - u² = 2ah

replacing the variables

v² - (0)² = 2(10)(63)

v² = 1260

v = √1260

v = 35.5 m/s

mistakes are ok

f=ma newton's second law 7x8=56n