Question

A ball of mass 200 g falls from a height of 2 m . What is the kinetic energy during the ball when it reaches the ground.

Answer 1

loss in potential energy = gain in kinetic energy 
m = 0.2 Kg.
g = 10 m/sec²
h = 2 m
loss in potential energy = mgh = 0.2*10*2 = 4 joule
hence kinetic energy of the ball when it reaches the ground = 4 joule 

Answer 2

Answer:

Explanation:

Solution :-

Given :-

Mass of the ball, m = 200 g = 200 × 10⁻³ kg

Height or Distance, d = 2.0 m

Acceleration due to gravity, g = 9.8 m/s²

To Calculate :-

K.E of the ball when it reaches the ground = ??

Formula to be used :-

P.E at the top = P.E at the ground

P.E = mgh

P.E = 200 × 10⁻³ × 9.8 × 2.0

P.E = 3.92 J

K.E of the ball when it reaches the ground its same as P.E at the top.

K.E = 3.92 J

Hence, the K.E of the ball when it reaches the ground is 3.92 J.