bolini
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Can anyone help me with the terminal velocity of a lacrosse ball?
Approximations would help too.
Thanks.
Approximations would help too.
Thanks.
The discussion revolves around calculating the terminal velocity of a lacrosse ball, including approximations and the necessary drop height to reach that velocity. Participants are exploring concepts from introductory physics, particularly related to motion and forces acting on falling objects.
The conversation is ongoing, with participants providing insights and asking for further clarification on incorporating air resistance into calculations. Some have shared their calculations and approximations, while others are questioning the assumptions made in the problem setup.
There are mentions of practical constraints, such as the height required to achieve terminal velocity and the complexity of modeling air resistance. Some participants express uncertainty about the derivation of the terminal velocity formula and the implications of varying air resistance during the fall.
bolini said:Not sure how to include resistance in the equation because it varies depending on the velocity which is changing until 33m/s is achieved.
Actually, an object never reaches terminal velocity, it only approaches it. The solution to the time and distance it takes to 'approach' terminal velocity (say reach .99V_t) involves a non linear differential equation (mg - cpAv^2 = mdv/dt), the solution of which is beyond me at this point, but it involves a natural log term I think and probably 'e' raised to some power. But on a practical level, terminal velocity is ordinarily reached (approached) in less than 9 seconds (but much less for lighter objects like a feather or penny). So you are correct in that it would take a height much greater than 180 feet for the ball to attain a speed of 33m/s or so when you consider air resistance. I'm guessing that it might be around 500 feet or so.bolini said:I'm attempting to help my son with a elementary school project that measures the impact crater of a ball at different heights. I thought it would be cool to see if we could achieve terminal velocity, but at 180 feet without considering air resistance, I think we are already too high to practically consider.
I believe the first equation basically used air resistance as the square of velocity. I am curious now, even though we probably won't look for a building or bridge high enough, what the height actually would be. If you could help me incorporate air resistance into the kinematic equation I used, that would be helpful.
Thanks for your help up until this point. I forgot how interesting physics used to be for me.