If a ball is thrown upward in presence of air resistance, would you expect the time during which it took to rise to be longer or shorter than the time it takes for it to fall?
Let's say it's a soccer ball

Question

If a ball is thrown upward in presence of air resistance, would you expect the time during which it took to rise to be longer or shorter than the time it takes for it to fall?
Let's say it's a soccer ball

happytalesrebornn · Answer

Of course.

anonymous · Answer

It is equal?

happytalesrebornn · Answer

Of course not. If upper-displacement force was applied to the ball against gravity /w air resistance, the ball will linger in the air longer than the fall.

anonymous · Answer

Hi Armi - what do you think the answer is ?

anonymous · Answer

Consider that the ball travels the same distance going up and coming down.
However, the deceleration going up is greater than the acceleration coming back down.
That means that at each point on the trajectory, the ball must have been moving faster on the way up, compared to the speed at the same position on the way down.

Conclusion - the time to go up is shorter than the time to come back down.

happytalesrebornn · Answer

Prof not necessarily.  Depends on initial force.

anonymous · Answer

What do you mean by initial force ? The ball rises and falls in the uniform gravitational pull at the surface of the earth, together with a drag force from the air. The forces are quite clear.

happytalesrebornn · Answer

Sigh, I'm done.  Continue as you may.

anonymous · Answer

My answer was the the two are the same: i tried to say that:
$$ma=mg+kv^2$$
so I tried to solve this differential equation:
$$dV/dt=g+k/m v^2$$

anonymous · Answer

And I got something like this:
|dw:1410817920895:dw|
which is some tangent function