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I am trying to set up this problem, but I am not sure how to go about doing so. (From University Physics, Young & Freedman):

You throw a baseball straight up. The drag force is proportional to $v^2$. In terms of $g$, what is the y-component of the ball’s acceleration when its speed is half its terminal speed and (a) it is moving up? (b) It is moving back down?

I am not sure if I am on the right track, but when they state that the drag force is proportional to $v^2$, it suggests to me that I need to use this relationship: $v^2 = \frac{mg}{D}$. Accordingly friction due to air drag is: $$f = Dv^2 = mg \rightarrow f = w$$

I am not sure how to find the y-component of the acceleration vector upwards or downwards at half the terminal speed - is the acceleration vector distinct from $g$ in this case?

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Have a look at physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf as this describes how to calculate the equations of motion including quadratic drag. –  John Rennie Oct 17 '12 at 7:31

1 Answer 1

Using the identity $F = ma = Dv^2 - mg$, I came up with the following solution.

Solving in terms of $v$, I should get $\sqrt{\frac{mg}{D}}$. Since I want to find the force of the y-component at half speed, I halve this value and get $\frac{\sqrt{\frac{mg}{D}}}{2}$. Plugging this back into the above equation for $v$, I get the following: $$ F = D(\frac{mg}{4D}) - mg \rightarrow F = \frac{mg}{4} - mg$$

I believe this will give $\frac{-3mg}{4}$ when the ball is falling, and I am assuming that it will be $\frac{5mg}{4}$ when going up.

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