Question

A body mass m is projected vertically downward from a point above the ground with a speed v.The kinetic energy of the body,just after falling through a vertical distance h is

Answer 1

Initial speed =$v$

Let the final speed be $v^{\prime }$

Vertical distance traveled =$h$

From kinematics,

$v^{\prime 2}-v^2=2gh$

$v^{\prime 2}=v^2+2gh$

Kinetic energy after falling a distance $h$:

KE=$\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m(v^2+2gh)=\frac{1}{2}m{v}^2+mgh$