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

  • #1
Dman0500
4
2
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?
 
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  • #2
What are forces on the rock?
 
  • #3
I suppose just g and a(y-axis)
 
  • #4
Dman0500 said:
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?
 
  • #5
Alright I get it now!
alright so when something is thrown upward, there is still only g?
 
  • #6
Dman0500 said:
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}##!
 
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  • #7
Make way more sense thank you
 
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Likes berkeman and etotheipi
  • #8
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}##.
 

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