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It can't be because of less surface area as we know that friction is independent of the surface area

Qmechanic
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1 Answers1

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It is because of the air resistance which heavily depends on velocity and geometry of the object.

Imagine a race car and a truck - which one gets more air resistance for the same speed?

The same argument applies to jumping out of a plane with or without a parachute. More surface area means more air resistance which equals lower (terminal) velocity.

If it was not for the air resistance, the terminal velocity would depend only on the altitude. The terminal velocity is a velocity at which object hits the ground.


Although this is not directly related to your question, velocity of the object also plays important role in calculation of the air resistance. In general, the force is proportional to velocity squared $F \propto v^2$.

Imagine two balls of the same size thrown from the same altitude at the same time, but one ball is 10 times heavier than the other. Which one will hit the ground first? According to the experiment done by Galileo in Pisa, we expect that they will hit the ground at the same time. However, the heavier ball can take more air resistance due to larger gravity, which means higher velocity at which air resistance and gravity are equal, i.e. heavier ball will hit the ground first.

Equation that describes free-fall with air resistance is:

$$ma = mg - F_\text{air}, \qquad F_\text{air} \propto v^2, \qquad a(t) = \frac{d}{dt}v(t)$$

where $v$ is the velocity at which the object falls and $a$ is the velocity rate of change (acceleration). It is obvious that velocity will be increasing as long acceleration is positive ($a > 0$), i.e. as long gravity is greater than air resistance ($mg > F_\text{air}$). However, the acceleration is not constant as predicted by Galileo, but it decreases towards zero due to the increasing $F_\text{air}$. This means that fall velocity increases less and less until acceleration is zero ($mg = F_\text{air}$), at which point velocity stops increasing and remains constant. This velocity is known as the terminal velocity.

Marko Gulin
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    I like your answer (+1) but you should elucidate the concept of terminal velocity, which I feel is alien to the OP. – Gert Dec 07 '21 at 16:57