Question

A body of 2kg falls from rest . What will be its kinetic energy during the fall at the end of 2s?

Answer 1

the answer is
1/2×2×2
cancel 2 by 2
So you have 2
than square it
you have 4

Answer 2

As acceleration is constant we can use the equation, v = u + at where u is the initial velocity, a is acceleration and v is the velocity after a time t.
Initial velocity, u = 0
Acceleration = acceleration due to gravity,g = 9.8 $ms^{-2}$.
time = 2s.
Therefore, v = 0 + 9.8 x 2 = 19.6 m/s.
Given that mass is 2 kg.
So kinetic energy at that instant, KE = $\frac{1}{2}mv^2$
Then, KE = $\frac{1}{2}\times 2\times (19.6)^2 =$ =384.16 J.