Question

A body of mass 2 kg falls from rest. What will be its kinetic energy during the fall at the end of 2 s?(assume g=10 m/s2)

Answer 1

u=0
m=2 kg
first we have to find v:
wkt, 
v=ut+1/2 at^2
v=0+1/2 *10*4
v=20m/s.

Now substitute this in KE equation as folows:

KE=1/2mv^2
      = 1/2 *2 *400
      = 400 joule.

Answer 2

given,
mass of the body ,m =2kg
initial velocity (u) =Om/s
time (t) = 2s
Acceleration due to gravity (g) =10m/s square
we need to find ,final velocity (v) =?
we know that
v=u+at
v= 0+10 × 2
v= 20 m/ s
K.E =1/2(mv square )
K.E =1/2 ×2×(20) square
K.E= 400 j
Hence , the kinetic energy of the body during the fall at the end 2s= 400 j


I hope this helpful