A flowerpot falls off the top of a building which is 70 meters high. If the flowerpot has a mass of 2.4 kg, what is its potential energy after 2.5 seconds? Keep 3 significant figures.
Height of the building = 70 meter
Mass of flowerpot = 2.4 kg
Potential energy after 2.5 seconds of free fall = “?”
By Newton’s Second Law of motion
we have: s = ut + 1/2at^2
In this case,
u=0, t=2.5 sec., a = g(9.8m/s^2).
So the vertical distance travelled by the free falling flowerpot, in 2.5 seconds
i.e. Height of flowerpot after 2.5 seconds = (70 - 30.625) meter
Therefore, potential energy after 2.5 seconds of free fall