Answer:
C. They are traveling at the same speed.
Explanation:
Lets take height of the building = h m
The mass of the objects = m kg
Initial velocity = u m/s
When objects thrown upward :
By using energy conservation
Work done by all the forces =Change in the kinetic energy of the objects
[tex]m g (-y) + m g ( h + y) = \dfrac{1}{2}mv^2- \dfrac{1}{2}mu^2[/tex]
[tex] mgh= \dfrac{1}{2}v^2- \dfrac{1}{2}u^2[/tex]
[tex]v=\sqrt{2gh+u^2}}\ m/s[/tex]
When objects thrown downward :
By using energy conservation
Work done by all the forces =Change in the kinetic energy of the objects
[tex] mg h= \dfrac{1}{2}mv^2- \dfrac{1}{2}mu^2[/tex]
[tex]mgh = \dfrac{1}{2}v^2- \dfrac{1}{2}u^2[/tex]
[tex]v=\sqrt{2gh+u^2}}\ m/s[/tex]
We can say that both the objects are having same velocity when they hit the ground.
Therefore the answer will be C.