Answer:
They both have the same speed when hitting the ground below
Explanation:
Conservation of the Mechanical Energy
In the absence of external non-conservative forces, the total mechanical energy of a particle is conserved or is kept constant.
The mechanical energy is the sum of the potential gravitational and kinetic energies, i.e.
[tex]\displaystyle M=mgh+\frac{mv^2}{2}[/tex]
When the first ball is launched at the top of the building of height h, at a speed vo, the mechanical energy is
[tex]\displaystyle M=mgh+\frac{mv_o^2}{2}[/tex]
When the ball reaches ground level (h=0) the mechanical energy is
[tex]\displaystyle M'=\frac{mv_f^2}{2}[/tex]
Equating M=M'
[tex]\displaystyle mgh+\frac{mv_o^2}{2}=\frac{mv_f^2}{2}[/tex]
We could solve the above equation for vf but it's not necessary because we have derived this relation regardless of the direction of the initial speed. It doesn't matter if it's launched with an angle above the horizontal, directly horizontal, or even directly downwards, the final speed is always the same.
It can also be proven with the exclusive use of the kinematic equations.
Note: The speed is the same for both balls, but not the velocity since the direction of the final velocity will be different in each case. Its magnitude is the same for all cases.
Answer: They both have the same speed when hitting the ground below
