A person standing near the edge of the top of a building throws two balls A and B. the ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed v_{A} and the ball B this the ground with a speed v_{B}. We have

#### Options

*v*_{A}>*v*_{B}*v*_{A}<*v*_{B}*v*_{A}=*v*_{B}the relation between v

_{A}and v_{B}depends on height of the building above the ground.

#### Solution

*v*_{A} = *v*_{B}

Total energy of any particle = \[\frac{1}{2}m v^2 + mgh\]

Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.

At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.