Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Have a look at as this describes how to calculate the equations of motion including quadratic drag. – John Rennie Oct 17 '12 at 7:31

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.