Question

A body of mass 2 kg is dropped from a building of height 5 m. The maximum kinetic energy the body can attain is ___ J.

Answer 1

From $1^{st}$ equation of motion we have, $v^2=u^2+2gh$

As it is dropped so, $u=0$ and $g=10m/s^2$ and $h=5m$

Putting these values we get, $v=10m/s^2$

So, the body would attain the maximum kinetic energy when it strikes the ground, i.e. K.E = $\frac{1}{2}\times m\times v^2=\frac{1}{2}\times 2\times {10}^2=100J$

Alternate way:

According to the law of conservation of energy, during free fall, the potential energy of the body at maximum height will be equal to the maximum K.E of the body. i.e. maximum potential energy=maximum kinetic energy.

So maximum potential energy $=mgh=2\times 10\times 5=100J$ which is equal to kinetic energy.