Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 would like to know how much speed is loss when a baseball is pitched. When a pitcher hurls the ball, it must lose speed before the catcher catches it because of the air resistance. So, if the catcher catches a 90 mph fastball, what speed was it going when it left the pitchers hand? The drag coefficient for a ball is 0.47 and the distance traveled is 60.5 feet. Let's make it easier by ignoring gravity.

How do I figure this out?

share|cite|improve this question

closed as off-topic by David Z Apr 2 '14 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

This is not a homework question. I would like to know how to include drag for a moving object. As the speed decreases, the magnitude of the drag also decreases. This necessitates that the result depends on integrating the equation, but I'm not certain how to do this. – LDC3 Apr 2 '14 at 1:18
60.5 feet is the distance from the rubber to the plate, but pitchers release the ball between 54 and 55 feet from the plate. How to calculte the effect of drag on a pitched baseball is explained here: – DavePhD Apr 2 '14 at 1:41
The most accurate way to do this would probably be to look at actual footage of baseball, and review it frame by frame to determine the velocity. Alternately, if you want to numerically approximate it, you can just numerically solve $m \frac{d^2x}{dt^2}=-\frac{1}{2} A \rho C_D\left(\frac{dx}{dt}\right)^2$ backwards in time, i.e., tell the solver the final position and final velocity, and make it compute backwards. – DumpsterDoofus Apr 2 '14 at 1:45
Actually, I want to solve this:… – LDC3 Apr 2 '14 at 1:46
@LDC3: I posted an answer to your link. – DumpsterDoofus Apr 2 '14 at 2:29