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Suppose I have a projectile motion without friction, the minimum velocity is always at the top.

Now I read that with friction the minimum velocity is not always at the top? Is this true? If yes, why?

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  • $\begingroup$ By friction, do you mean something like air resistance? $\endgroup$ – Steeven Aug 16 '17 at 17:53
  • $\begingroup$ Yes, thats what I meant $\endgroup$ – Ayoub Rossi Aug 16 '17 at 17:59
  • $\begingroup$ I think it's not true. Even with friction like air resistance, the minimum velocity should always be at the top. $\endgroup$ – Wrichik Basu Aug 16 '17 at 18:02
  • $\begingroup$ @WrichikBasu I don't think it's always the case when you add in friction. If the drag force in the x direction were greater than gravity, right after the apex, the magnitude of velocity may be lower than at the apex, because horizontal velocity decreases more than vertical velocity increases. $\endgroup$ – JMac Aug 16 '17 at 18:05
  • $\begingroup$ If I throw a rock in a pond it may go faster at the top than when it is sinking in the water. $\endgroup$ – Keith McClary Aug 16 '17 at 18:42
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This can certainly be the case. Take for example an imaginary catapult. Let's fire a round rock from it. As long as we're not using an exceptionally powerful catapult, the trajectory will be close enough to the one without air resistance that the minimum velocity will be at the top of the arc, very near what it would be without air resistance.

Now imagine you use the same catapult and fire a wad of paper. This paper wad will not have enough inertia to maintain its velocity and will slow down considerably on the way up. So much so that it will deviate considerably from its theoretical trajectory and reach its maximum height very early. We will assume that it has not yet reached its terminal velocity. From here it will accelerate towards the ground until it reaches its vertical component reaches terminal velocity. If you are very lucky, it will reach that terminal velocity in the vertical component before it loses all of the horizontal component of its velocity. If this is the case, minimum speed will be reached when all horizontal speed is lost.

Alternatively you could consider a suborbital trajectory where it reaches its maximum height outside the atmosphere and then is slowed as it reenters, but I feel that was probably not in the spirit of the question.


Update from comment:

This is also possible when the projectile's horizontal acceleration is higher than its vertical acceleration. This can create a situation where, just after the apex, the projectile has lost more velocity in the horizontal direction than it has gained in the vertical direction. The minimum speed in this case could occur here, and could be slower than the terminal velocity.

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    $\begingroup$ The terminal velocity point is a good one. You can also consider that if the horizontal drag force is greater than gravitational force, the decrease in horizontal velocity will be greater than the increase in vertical velocity at the apex. This would mean that immediately after reaching the top, it will have to be slower. So basically any big enough light enough object with enough horizontal velocity will do this. $\endgroup$ – JMac Aug 16 '17 at 18:45
  • $\begingroup$ Good point. I'll add it to my answer. $\endgroup$ – bendl Aug 16 '17 at 18:46
  • $\begingroup$ Is it really possible? Could the air break the paper ball more strongly in the way up than in the way down? $\endgroup$ – stafusa Aug 16 '17 at 18:59
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    $\begingroup$ Air resistance is a function of velocity, so it will absolutely slow down more on the way up than the way down. As for whether or not it's possible, I just said it was ;) In any case, it's not so much about where the air resistance is strongest, but more about where the apex is in relation to it's velocity. $\endgroup$ – bendl Aug 16 '17 at 19:02
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    $\begingroup$ In fact, I might venture to suggest (without any sort of proof mind you) that it would actually be uncommon to achieve minimum speed at the apex for a sufficiently light projectile with plenty of air resistance. $\endgroup$ – bendl Aug 16 '17 at 19:04
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There is a quite simple way to understand this. That the minimum velocity isn't reached at the top is pretty obvious for a projectile which is fired at an angle close to horizontal. Suppose its initial velocity is much higher than the terminal velocity that air friction imposes as it falls. The projectile will at some point reach its terminal velocity, which is smaller than its initial velocity. Since the angle at which the projectile was fired was small, it will reach the top of its trajectory pretty soon. In particular, it will reach the top before reaching terminal velocity. It is thus clear that the minimum velocity will be reached later, at some point along its trajectory as the projectile falls towards the ground (and once it loses its horizontal velocity).

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Gravity and drag both reduce speed $v$ on the upward section of the trajectory, so speed decreases monotonically to the highest point (the apex). It follows that the minimum speed cannot occur before the apex.

On the downward section the vertical component of speed $v_y$ increases monotonically toward the terminal speed $v_t$, while drag decreases the horizontal component $v_x$ monotonically towards zero. If the speed at the apex is $v_1 \gt v_t$ then it is obvious that the minimum speed does occur after the apex (ie not at the top), because it must decrease to $v_t$ eventually. However, it is not so obvious whether $v$ falls below $v_t$ and then increases back to $v_t$. If $v_1 \lt v_t$ then it is not obvious or intuitive whether $v$ increases monotonically to $v_t$ after the apex or dips to a minimum. Mathematical analysis is required.

In the article Minimum and Terminal Velocities in Projectile Motion the authors report the surprising result that for drag of the form $kv^n$ the minimum speed is always lower than both terminal speed $v_t$ and the speed $v_1$ at the apex :

At the beginning, the effect of the drag is stronger than the effect of gravity if the initial speed is large enough. But once the velocity has diminished, gravity becomes dominant and accelerates the projectile until the terminal velocity is reached. There is an interplay between two opposite causes (drag and gravity) that produces the curious behavior shown in Figure 3....

Naively, one would have expected that the minimum velocity is $v_t$, but it comes out that the projectile speed diminishes until it reaches a minimum and then it starts increasing. $v_t$ is always reached from below. This means that the velocity is not a monotonous function but it shows a minimum different from both $v_1$ and $v_t$. This behavior is the same for any value of the power n in the drag force.

enter image description here

The figure shows what happens with $V=v_1/v_t=0.95$ and $1.05$, for the case n=3. Angle $\theta$ is the inclination of the trajectory below the horizontal.

The authors also consider a projectile launched with speed $v_0$ at an angle above the horizontal. They find that if $v_0 > v_t$ then the minimum speed is $v_t$ for initial angles above a critical value $\theta_1 \approx 50^{\circ}$. For $v_0 < v_t$ there is a another critical angle $\theta_2 \approx 10^{\circ}$ above which the minimum speed is the launch speed $v_0$. Speed is a monotonous function above those critical angles; below them, speed is non-monotonous and the minimum is lower than both the launch speed $v_0$ and the terminal speed $v_t$. [The authors' results in this paragraph seem to be erroneous, because the minimum speed cannot occur before the apex.]

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