Dynamic Equilibrium -- Acceleration of a rock thrown from a bridge

[Dman0500]

Homework Statement:

A rock is thrown from a bridge at an angle 30° below horizontal. Immediately after the rock is released, is the magnitude of its acceleration greater than, less than, or equal to g? Explain.

Relevant Equations:

Conceptual

I know the acceleration of the rock is equal to g, but why. If we neglect air resistance, what is actually making the rock fall? Wouldn't it be that g overcomes the acceleration of the y plane at some point so the rock starts coming down or in this case accelerate more by throwing below 0 degrees?

 

What are forces on the rock?

 

[Dman0500]

I suppose just g and a(y-axis)

 

I suppose just g and a(y-axis)

There is a weight force, of magnitude $mg$, acting downward. That is the only force! "$a$" is not a force, it is an acceleration!

What is Newton's second law?

 

[Dman0500]

Alright I get it now!
alright so when something is thrown upward, there is still only g?

 

Alright I get it now!
alright so when something is thrown upward, there is still only g?

Yes, something undergoing free-fall (only force is the gravitational $m\vec{g}$) will accelerate at $\vec{g}$. Make sure you are clear to distinguish forces from accelerations!

The result follows from Newton's first law; the net force is $m\vec{g}$, and Newton tells us that $m\vec{g} = m\vec{a}$. So $\vec{a} = \vec{g}$.

If you throw something downward, it has an initial downward component of velocity. However the acceleration is still just $\vec{g}$!

 

[Dman0500]

Make way more sense thank you

 

No problem. If you are ever in doubt about similar matters, draw a diagram! Draw the forces on the diagram, and then use $\vec{F} = m\vec{a}$.