Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value 10 m/s². The work done by the (i) gravitational force and the (ii) resistive force of air is : 

i) 10 J  ii) -8.75 J

According to work energy theorem

w_g + w_R = \(\frac{1}{2}mv^2\)

w_g = mgh

     =10^-3 x 10 x 10³

     = 10J

 10 +  w_R =\(\frac{1}{2}\) x 10^-3 x 50 x 50

 10 +  w_R = 1.25

w_R = -8.75J