A body of mass $2 \,kg$ is projected vertically upwards with a velocity of $2\,m\,{\sec ^{ - 1}}$. The K.E. of the body just before striking the ground is ............. $\mathrm{J}$
A body of mass $2 \,kg$ is projected vertically upwards with a velocity of $2\,m\,{\sec ^{ - 1}}$. The K.E. of the body just before striking the ground is ............. $\mathrm{J}$
$2$
$1$
$4$
$8$
A body of mass $2 \,kg$ is projected vertically upwards with a velocity of $2\,m\,{\sec ^{ - 1}}$. The K.E. of the body just before striking the ground is ............. $\mathrm{J}$
If 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} = 4\,J$