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CONTEXT:

When there is no air resistance, the time taken for a particle to go up $T_U$ will of course be equal to the time taken for the same particle to fall down $T_D$. So in this case, we have $$T_U = T_D$$

But when there is a resistive force say $mkv$ or $mkv^2$, where $m$ is the mass of the particle, $k$ is a positive constant and $v$ is the velocity of the particle, we no longer have such nice symmetry properties. Hence $$T_U \neq T_D$$ So it is only natural to wonder which of the two is larger, and to prove it analytically.

THEIR METHOD:

I have found THREE sources that use the same technique to prove that $T_D > T_U$. However, I think the technique used may not be correct.

The technique follows these steps, roughly speaking. I'll omit the details for now, as my question is regarding the validity of the technique, rather than the computations.

  1. Form the acceleration-velocity separable differential equation.
  2. Find velocity as a function of time using $a=\frac{dv}{dt}$.
  3. Re-arrange and make $v$ the subject, then use the fact that $v=\frac{dx}{dt}$ to find a displacement-time equation $x(t)$.
  4. Find the time $T_U$ taken to reach maximum height.
  5. Find $x(2T_U)$ and observe that $x(2T_U)>0$. Hence, the downwards journey takes longer.

MY ISSUE:

In the above technique, we are using the same $x(t)$ function, which models upwards motion, to model the downwards motion too.

But I do not think this is valid because in the upwards motion, we have two forces working against us: weight force and resistive force. However, when we are moving downwards, the weight force is now assisting the particle's motion. In other words, the $x(t)$ function will no longer be valid if $t>T_U$.

I cannot see any reason why this technique should work, yet I've found three sources all doing the same thing, so I am inclined to believe that I am missing something here, which is why I come to Physics SE. Could somebody shed some light on this scenario?

SOURCE:

I am very uncertain about the line

For $t \geq t_m$, $v<0$ and therefore $f(v)<0$.

because $f(v)=v^2$ gives an easy counter example to this claim.

enter image description here

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  • $\begingroup$ How did you conclude that $x(t)$ only models the upward motion? The same forces act on the object throughout the flight - just one of them depends on velocity and changes sign when velocity changes sign, but $F=ma$ is valid for both upwards and downwards motion, and so is its solution. $\endgroup$
    – ACuriousMind
    Jul 31, 2016 at 16:08
  • $\begingroup$ I understand your point. This would make sense if the resistive force was $mkv$ so that the sign changes when the velocity also changes sign. But what about if the resisitve force was $mkv^2$? In this case, the technique is invalid. $\endgroup$
    – Trogdor
    Jul 31, 2016 at 16:11
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    $\begingroup$ A force is a vector. $mkv^2$ would only give the magnitude of that vector - but what determines its direction? $\endgroup$
    – ACuriousMind
    Jul 31, 2016 at 16:16
  • $\begingroup$ In that case, the sign in front would determine the direction. This is where I have issue. For $mkv^2$ we will need to consider two differential equations. The first is $a=g-kv^2$ for when the particle falls down, and the other is $a=-g-kv^2$ for when the particle goes up. I do not think using only $a=-g-kv^2$ is valid for both up and downwards motion, which is implied in one of the sources. I will update my main post with the problematic line for reference. $\endgroup$
    – Trogdor
    Jul 31, 2016 at 16:22
  • $\begingroup$ You haven't named the source and not given their definition of $f$, but if they say that $f(v) < 0$ if $v < 0$, then that belongs to the assumptions. Which are obvious if $f$ is (the negative of) a friction force, because friction by definition acts opposite to the direction of movement. $\endgroup$
    – ACuriousMind
    Jul 31, 2016 at 16:34

2 Answers 2

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You seem to agree that $T_D > T_U$ as the sources prove. Your only concern is the validity of using the same notation $x(t)$ for both the up and down motions.

My interpretation is that $x(t)$ is a piece-wise function. The sources are not assuming that it has the same form going up ($t<t_m$) as coming down ($t>t_m$). Your argument seems to assume that $x(t)$ will have the same form in both cases. It is this assumption which is wrong. I do not see anything wrong with their method.

I think ACuriousMind and Trogdor are right : the magnitude of the resistance force is $f(v)=mkv^2$. The direction of that force is indicated by the sign of $f(v)$.

Your own method of solution (which presumably runs as follows) seems perfectly adequate to me without any need to carry out the integration. I don't understand why you are quibbling about the analytic approach.


Non-Analytic Solution

On the way up gravity and drag act in the same direction, whereas on the way down they act in the opposite direction. This means that, at each height, the deceleration on the way up is always greater than the acceleration on the way down. It follows that, at each height, the speed on the way up will also be greater than the speed on the way down, and therefore the time to cover the same distance going up will always be smaller than the time coming down.

Graphically, plot distance $x$ downwards from the highest point vs time $t$ for up and down motions on the same axes. To do this, run the up-motion backwards in time. The up-motion accelerates more rapidly than down-motion, so the $x(t)$ graph curves upwards more steeply for the up-motion than for the down-motion. Therefore a given distance $x$ is reached in a shorter time for the up-motion.

This is an application of the Racetrack Principle : if $f'(t)>g'(t)$ for all $t>0$ and $f(0)=g(0)$ then $f(t)>g(t)$ for all $t>0$.

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On the way up gravity and drag act in the same direction, whereas on the way down they act in the opposite direction. This means that, at each height, the deceleration on the way up is always greater than the acceleration on the way down. It follows that, at each height, the speed on the way up will also be greater than the speed on the way down, and therefore the time to cover the same distance going up will always be smaller than the time coming down.

Graphically, plot distance x downwards from the highest point vs time t for up and down motions on the same axes. To do this, run the up-motion backwards in time. The up-motion accelerates more rapidly than down-motion, so the x(t) graph curv

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