King Kong carries his chapstick up the 321m tall Empire State Building. At the top of the building, he drops it (oh no!). How fast will the chapstick be moving when it hits the ground?
We need this formula
vf^2 = vi^2 + 2 (a)(d)
vi is the initial velocity = 0
a = acceleration due to gravity = -9/81m/s^2
d = the displacement = 321 m
vf = the final velocity ......so we have....
vf^2 = 0 + 2 (-9.81m/s^2)(321m)
vf^2 = 6298.02 m^2/s^2 take the square root
vf ≈ 79.36 m/s