Question

A water drop of mass 2.0 g starts falling from a height 2.0 km. It hits the ground with speed 50.0 m s1. The work done by the resistive forces of air during flight of the drop is (g = 10 m s2)

Answer 1

Answer:

total work done =change in kinetic energy

Explanation:

mgh +x =kf-ki

2#10*-3#2#10*3 + x =1/2#2#10*3+0