1
$\begingroup$

A ball of mass M is thrown vertically upward with an initial speed of v o. It experiences a force of air resistance given by F = -kv, where k is a positive constant. The positive direction for all vector quantities is upward. Express all algebraic answers in terms of M, k, v o , and fundamental constants.

The solution says: "It takes longer for the ball to fall. Friction is acting on the ball on the way up and on the way down, where it begins from rest. This means the average speed is greater on the way up than on the way down. Since the distance traveled is the same, the time must be longer on the way down."

I don't understand how they can claim the average speed on its way up is greater than on its way down.

Can someone clarify this with great detail? Thank you in advance...

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ This question is similar to: Time of a ball going up and down with air resistance. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$
    – BowlOfRed
    Commented Sep 16 at 7:54

2 Answers 2

Reset to default
2
$\begingroup$

I don't understand how they can claim the average speed on its way up is greater than on its way down.

Because the total energy of the ball is smaller on its way down because of air drag, but for each height the potential energy is the same on the way up and the way down. Therefore, at each height the kinetic energy (and, therefore, the speed) is smaller on the way down, so the average speed of the ball is smaller on its way down.

Improve this answer
$\endgroup$
7
  • $\begingroup$ How do you know the kinetic energy is smaller on its way down at any given height? $\endgroup$
    – Andrew Steane
    Commented Sep 17 at 8:06
  • $\begingroup$ @AndrewSteane : The only forces acting on the ball are gravity and air drag. Gravity is conservative, so it does not change the total energy. Air drag is nonconservative and performs negative work, so it reduces total mechanical energy of the ball. Thus, the total mechanical energy at the same height is lower on the way down than on the way up, whereas the potential energy is the same. So the kinetic energy is lower at the same height on the way down than on the way up. $\endgroup$
    – akhmeteli
    Commented Sep 17 at 13:43
  • $\begingroup$ But surely this argument applies equally on the way up as on the way down. $\endgroup$
    – Andrew Steane
    Commented Sep 17 at 13:46
  • $\begingroup$ but air drag also acts on the ball while going up equally? $\endgroup$
    – Amit Verma
    Commented Sep 17 at 14:01
  • 1
    $\begingroup$ @AndrewSteane : Yes, but the ball goes down AFTER it goes up. $\endgroup$
    – akhmeteli
    Commented Sep 17 at 20:13
0
$\begingroup$

There are two forces acting: gravity and air resistance. On the way up these act in the same direction; on the way down they act in opposing directions. Therefore the magnitude of the acceleration is larger on the way up than on the way down. For simplicity let's imagine the acceleration as constant in each part of the journey (it isn't constant but this approximation should be good enough to treat the kind of overall average the question is asking about.) Let's plot a graph of $v(t)$. It starts at some positive $v$ at $t=0$, falls to $v=0$ at some moment $t_{\rm top}$, and then continues to fall to some final $v$ which will be negative. We know the following:

  1. The slope of the graph is steeper on the way up (i.e. for $t < t_{\rm top}$).
  2. The area between the line and the horizontal axis is the same in the two parts of the graph.

If you draw such a graph using straight lines you will see it looks like two triangles. In order to have the same area, the horizontal width of the second triangle has to be longer than the horizontal width of the first one. Therefore the balls takes longer to come down.

Even after deriving this I still find it counter-intuitive.

Improve this answer
$\endgroup$

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