Question

A body of mass $2$ kg is projected vertically upwards with a velocity of $2$m $se{c}^{-1}$. The K.E. of the body just before striking the ground is?

• A. $2$ J
• B. $1$ J
• C. $4$ J
• D. $8$ J

Answer 1

The correct option is C $4$ J

Applying energy conservation at time of projection, and returning at ground.

$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2$

$⟹u=v$

It shows that particle is projected vertically upward with velocity of $2m/s$ then it returns with the same velocity.

So its kinetic energy=$\frac{1}{2}m{v}^2=\frac{1}{2}\times 2\times (2{)}^2=4J$