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.
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.