Question

a ball of mass 200 g falls from a height of 2.0 m . what is the kinetic energy of the ball when it reaches the ground ? take g = 9.8 m/s^2

Answer 1

m = 200 g = 200/1000 kg = 200 * 10^-3 
h = 2 m
 g = 9.8 m/s
to calculate k.e. of the ball reaches the ground = ?

P.E. at the top = P.E. at the ground 
                        = mgh 
                         = ( 200 * 10^-3 )9.8 * 2
                         = 3.92 joules
therefore , K.E. of the ball when it reaches the ground is same as P.E. at the top
                          => 3.92 j  

Answer 2

Given - 

Mass of ball = 200 g = $\frac{200}{1000} kg$ = 0.2 kg

Height at which the ball was = 2 m

g = 9.8 m/s

According to the law of conservation of energy,

P.E of the ball = K.E of the same ball

mgh = K.E of the ball

(0.2 * 9.8 * 2) = K.E of the ball

3.92 J = K.E of the ball