0
$\begingroup$

For example, a body in free fall on earth and the same body experiencing the same acceleration ($ g $ in this case) but horizontally, would have the same terminal velocity? (considering that the contact area is the same in both cases)

$\endgroup$
2
$\begingroup$

It depends on what you mean by "direction of movement".

If you mean the direction that an object is moving at any given moment, no, because terminal velocity is a hypothetical value that doesn't care for current circumstance.

If you mean the direction of movement for the object when it's at terminal velocity, then, yes. For example, a rocket with an engine going can have a terminal velocity in the upward direction or the downward direction, but its terminal velocity downward would tend to be faster than upward due to gravity.

$\endgroup$
  • $\begingroup$ Sorry to comment on the post here, but can you please be more careful in reviewing suggested edits? For instance, this edit you approved does not fully address the issues & actually incorrectly edits a word. $\endgroup$ – Kyle Kanos Jun 7 '18 at 14:02
  • $\begingroup$ @KyleKanos It's all good, though that edit does look like an improvement to me, albeit not a complete or perfect one. By the way, why'd you convert $\frac{x}{y} \to x/y$ for two of the fractions? $\endgroup$ – Nat Jun 7 '18 at 14:19
  • $\begingroup$ well the suggested edit changes the meaning of the sentence, which should be rejected in that grounds alone. I used x/y because \frac should never be used for inline equations. $\endgroup$ – Kyle Kanos Jun 7 '18 at 14:22
  • $\begingroup$ @KyleKanos That's probably a difference in perception; I didn't see the changed word as changing the sentence because the sentence already looked like gibberish, such that I saw that particular modification as lateral. This is, I wasn't sure what "aftec" was supposed to be, but since "after" looked like a reasonable enough guess (plus, it was an actual word), I overlooked it for what I saw to be the main improvement of the suggested edit, which would be the typos just before that. $\endgroup$ – Nat Jun 7 '18 at 14:32
  • $\begingroup$ I think then if it doesn't make sense one way or the other, it's better to reject it & leave a comment to the poster asking for clarification. $\endgroup$ – Kyle Kanos Jun 7 '18 at 14:38
2
$\begingroup$

Not necessarily. The atmospheric density has to be considered. If the object is falling downward, the atmospheric dehsity increases and that will increase the drag, If the object is going horizontally the atmospheric density is basically constant so drag will be constant.

If this is for a baseball, for example, the atmospheric pressure may be a constant throughout its flight due to the short distances involved.

It this is for a much faster object the atmospheric pressure is not constant. Also, the drag dependance on velocity is not a constant so that must be taken into account as well.

$\endgroup$
2
$\begingroup$

Yes. Air resistance is independent of direction of motion. It is always opposite to the direction of motion, it increases with speed and air density, and depends on the cross-sectional area and longitudinal shape of the object.

Terminal velocity occurs when air resistance balances resultant driving force (eg thrust or weight) in both direction and magnitude. If the resultant driving force acting on the body is the same, and all other conditions are the same, then the terminal velocity will be the same also.

$\endgroup$
1
$\begingroup$

Normally not. Below are measurements on an average human in the wind tunnel at different orientations. The values given are the drag force divided by the dynamic pressure; that's why they have the dimension of an area.

Drag on human body

This figure is copied from chapter 5 of Sighardt Hoerner's book "Fluid Dynamic Drag", 1965 edition.

Now consider that terminal velocity changes with the square root of the dynamic pressure and air density, so the speed differences between the different postures are as large as 1 : 2.74.

If the frontal area is the same, just consider a semisphere: With the closed side facing forward, its drag coefficient is 0.38 while the same sphere, now facing forward with the open side, will have a drag coefficient of 1.42. Consequently, the terminal speed difference between both orientations will be 1 : 1.93.

This does not consider dynamic stability, because such a semisphere will tumble while in free fall.

$\endgroup$
0
$\begingroup$

$$\vec{F}_\text{resistance}=-b\vec{v}$$ (assuming linear relationship between resistance and velocity) $$\vec{v}=\vec{v_x}+\vec{v_y}$$ $$v=\sqrt{v_x^2+v_y^2}$$ $$\frac{dv_x}{dt}=-\frac{b}{m}v_x$$ $$\frac{dv_y}{dt}=-g-\frac{b}{m}v_y$$ These differential equations do not have analytical solutions. They have to be integrated numerically. In the first case you have to set $v_x(t=0)=0$. In the second one you have to set $v_x(t=0)=u$.

$\endgroup$
  • $\begingroup$ The equations you wrote down seem to be linear and uncoupled. So I think that they can be integrated analytically. $\endgroup$ – nicoguaro Jun 4 '18 at 4:59

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