1. I have an arbitrary object with initial velocity $v_1$ pointing upwards. There's also air resistance proportional to the square of speed. So this object goes up, turns around, goes back down and lands with velocity $v_2$, happens to be smaller than $v_1$. What does that tell us about the time of up-motion? Is it smaller, even or greater than time of falling down?

I could probably solve numerically differential equation and somehow deduce it, but I'd rather see a reasoning which does not require use of computer.

  1. Bonus question: what happens in simplified case - when air resistance depends linearly on speed, not its square?

I don't necessarily want a full solution, but if someone wants to steal the show be my guest.


1 Answer 1


Air resistance, while dependent on velocity also only affects velocity. If the initial speed of the object is greater than the final speed of landing, then the time of flight upwards will be less than that down, because no matter what the faster initial trajectory up will cover the same height in less time.

The situation described is analogous to throwing a air filled ball high up at a reasonable speed, then watching it float back down


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