Why is total system momentum zero when a ball falls to h/2?

[Husker70]

Homework Statement

A system initially consists of a ball of mass m at rest at a height h above the Earth and the Earth, of mass Me. When the ball has fallen a distance h/2, the total momentum of
the system is
A. 0
B. m(sqrrt)2gh
C. Me(sqrrt)2gh
D. Me + m(sqrrt)2gh
E. Me - m(sqrrt)2gh

Homework Equations

The Attempt at a Solution

I put B and missed it. I think the answer is C but not sure.
Any help understanding this would be appreciated.
Thanks,
Kevin

 

[asleight]

Homework Statement

A system initially consists of a ball of mass m at rest at a height h above the Earth and the Earth, of mass Me. When the ball has fallen a distance h/2, the total momentum of
the system is
A. 0
B. m(sqrrt)2gh
C. Me(sqrrt)2gh
D. Me + m(sqrrt)2gh
E. Me - m(sqrrt)2gh

Homework Equations

The Attempt at a Solution

I put B and missed it. I think the answer is C but not sure.
Any help understanding this would be appreciated.
Thanks,
Kevin

$$\textbf{v}_f^2=\textbf{v}_i^2+2\textbf{ad}$$

so, $$\textbf{v}=\sqrt{2\textbf{ad}}$$, and:

$$\textbf{p}=m\textbf{v}=m\sqrt{2\textbf{ad}}$$.

The total momentum of the system is the sum of momenta of the particles. The above momentum is that of the ball. The momentum of the Earth is given by:

$$M_e\textbf{v_e}=$$,

add them and that's your answer which isn't readily available, sadly. Maybe someone else has more of an intuitive approach.