**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.**

**Answer:**

Given,

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

= 1/2*9.8*2.5*2.5
=30.625 meter.
i.e. Height of flowerpot after 2.5 seconds = (70 - 30.625) meter
=39.375 meter
Therefore, potential energy after 2.5 seconds of free fall
= m*g

*h*

=2.49.8*39.675 N-m

=2.4

=926.100 N-m