Question

A body of mass 2kg is dropped from the top of a tower of height 50m .If the acceleration due to gravity is 10m\s^2, what will be it's kinetic energy at the end of 2s.

Answer 1

For a freely falling body,
v=gt

where g is acceleration due to gravity and t is time.
g=10 m/s²
t=2s

v=gt
So v=10*2
       = 20 m/s

(Thus in 2s body would travel a distance of 40 m from release point)
                                                 
Kinetic Energy K=1/2 mv²

m=2kg
v=20m/s

So K= 1/2 * 2 *20²
        = 1 * 400
        = 400 J

Thus the Kinetic Energy of the body would be 400 J at the end of 2s.

Answer 2

Explanation:

$v = u + gt \\ v = 0 + 10 \times 2 \\ v = 20m \\ \\ kinetic \: energy \: = \: 1by2 \times m \times v \: square \\ = 1by2 \times 2 \times 20 \times 20 \\ = 400j$