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If I have understood it correctly, air resistance is proportioal to the squared speed of an object, meaning that the faster the object is going the more affected by air resistance it is ( which must mean it slows down?) Does this mean that a slow moving object will technically fall faster than a fast falling object? Would this mean that the lower the initial velocity the faster the object will go and vice versa? I am really confused.

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  • $\begingroup$ What it means is that for most rigid objects, as the object falls through air,* its speed approaches a constant terminal velocity. [* Same principle works for objects falling through other fluids besides air.] $\endgroup$ Sep 26, 2022 at 16:13
  • $\begingroup$ "which must mean it slows down?" - no, it means that its speed increases more and more slowly (a falling object accelerates as it falls, but the acceleration drops off due to drag, as Solomon Slow described above.) $\endgroup$ Sep 26, 2022 at 19:14

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Velocity is not the only factor that determines the drag force. Area of cross section, for example, is another factor affecting the magnitude of drag.

If everything else is same except the velocity, drag will in general be higher for higher velocities. Because other things are same, higher drag will cause higher deceleration. So speed will reduce at faster rate when the speed is higher.

Things, however, can be more complicated than above. For example, drag is not always proportional to the square of speed. It is in general at relatively higher speeds. At relatively lower speeds, drag tends to be proportional to the speed.

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Does this mean that a slow moving object will technically fall faster than a fast falling object?

No, one body can experience bigger resistance but have still bigger velocity, it's just different values. Resistance of air is the force acting on a body, velocity is velocity. The force is responsible for changing of velocity, that means initially velocities can be random, but the body experiencing bigger resistance (minus weight force) will change it's velocity faster, before both velocities won't equalize. You have to also consider the weight force, acting against air resistance and it's also different for different bodies, but if you will compare the same bodies with equal weight, then the logic (bigger resistance -> faster speed reducing) is working.

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Does this mean that a slow moving object will technically fall faster than a fast falling object?

I'm assuming by "fall faster" you are referring the rate of loss of altitude, i.e., the downward acceleration of the object. The downward acceleration of each object will depend on the net force acting on each object. That, in turn, will depend on two forces: Gravity and Drag. In order to answer your question we need to limit the number of variables involved with these two forces.

First, in order for the force of gravity to be the same, the objects need to have the same mass $m$. We will make the further assumption that at the time of comparison each object is at the same height above, and close enough to, the surface of the earth so that the acceleration due to gravity is constant during the remainder of the fall and is 9.8 m/s$^{2}$

Next, since the drag force is proportional to the velocity at slow speeds and the velocity squared at high speeds, we will need to assume one or the other. In order to apply the following equation for the drag force we will assume the high speed model:

$$F_{D}=\frac{1}{2}\rho v^{2}C_{D}A$$

where

$\rho$ = the density of the air (which will be the same at the same height, which we stipulated above)

$v$ = the speed of the object

$C_D$ = the drag coefficient

$A$ = cross sectional area projected in the direction of motion

In order to compare the drag force on each based only on their speeds (how fast they are moving) we will assume $A$ and $C_D$ are the same for both objects.

The net downward force acting on each is then

$$F_{net}=mg-\frac{1}{2}\rho v^{2}C_{D}A$$

From the last equation we see that the net downward force acting on the slow moving object (having a lower value of $v$) is greater than the fast moving object (having a higher value of $v$). Then since the acceleration of each is $F_{net}/m$ and each has the same mass, the acceleration of the slower moving object is greater than the fast moving object. Finally, given enough time for both, the drag force on each object will equal the force of gravity for a net force of zero with each object attaining its constant terminal velocity.

Hope this helps.

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  • $\begingroup$ This helped a lot, thank you! $\endgroup$
    – Manar
    Sep 27, 2022 at 22:13

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