Drag Force

  • #1
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Homework Statement



A gun is fired straight up. Assume a quadratic drag force.
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(upward motion) [tex]v^2 = Ae^{-2kx} - \frac{g}{k}[/tex]
(downward motion) [tex]v^2 = g/k - Be^{2kx}[/tex]

Homework Equations





The Attempt at a Solution


I can easily solve the integrals unfortunately I am having difficulty setting equations up.

Upward:
ma = mg + cv^2

Unfortunately, to get the book's answer I would need to say ma = -mg - cv^2
And I am having a difficult time seeing why both terms are negative.

I draw the free body diagram. As soon as the bullet leaves the barrel, the acceleration is in the downward direction. (so mg is pos) and the drag also acts in the downward direction so I thought ma = mg + cv^2. Shouldn't mg and cv^2 both be positive because they are in the direction of the acceleration?

Does anyone know why I am getting different signs?
 

Answers and Replies

  • #2
Dale
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Upward:
ma = mg + cv^2

Unfortunately, to get the book's answer I would need to say ma = -mg - cv^2
And I am having a difficult time seeing why both terms are negative.

I draw the free body diagram. As soon as the bullet leaves the barrel, the acceleration is in the downward direction. (so mg is pos) and the drag also acts in the downward direction so I thought ma = mg + cv^2. Shouldn't mg and cv^2 both be positive because they are in the direction of the acceleration?

Does anyone know why I am getting different signs?
They just defined the positive direction as upwards where you defined it as downwards. There is nothing wrong with either choice, as long as you are consistent while working the problem.
 

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