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A ball of unit mass is dropped. How do I work out it's terminal velocity when the ball has air resistance proportional to the square of the velocity?

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Let the position of the ball be $y(t)$. According to Newton's second law $$y''= -k y'^2 $$

Plugging in $y'=p(y)$ the equation becomes $$pp'=-kp^2 $$ which can be solved for $p(y)$ by separation of variables.

Can you take it from here?

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