Question

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
(a) v_A > v_B
(b) v_A < v_B
(c) v_A = v_B
(d) the relation between v_A and v_B depends on height of the building above the ground.

Answer 1

(c) 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.