1
$\begingroup$

Why does the velocity of a skydiver become constant again after opening the parachute? If by net F= ma, we will have net F smaller than 0 as the air resistance is increased greatly and the weight remains unchanged. Why does the acceleration becomes 0 again after decelerating?

$\endgroup$
4
$\begingroup$

There are two forces acting on the skydiver.

  1. The force ($mg$) due to the weight of the skydiver pulling him towards earth.

  2. The drag force ($-kv$) offered by the air as resistance to his motion. This force is in opposite direction to his velocity i.e. upwards (therefore negative).

Let's take the downward direction to be positive. Then net force is

$$F = mg - kv$$

$k$ is just a viscosity constant.

Equate $F$ with $ma$:

$$ma = mg - kv$$

Now you can intuitively see, that $mg$ is just a constant acceleration in the beginning. But as the parachute tries to speed up, the $kv$ factor increases in magnitude until it just equals $mg$. At that point, acceleration becomes zero. Now onwards the velocity will not change (zero acceleration).

This value of velocity is called terminal velocity.

PS: Have a look at the very relevant comments made by some people about what exactly is the effect of the parachute opening. In brief: The above discussion is valid whether the parachute is open or not(every object traveling through a viscous fluid experiences drag). The effect of opening the parachute will only increase the magnitude of drag ie the constant k(which is a function of the size and shape of the body) will change.

$\endgroup$
  • 2
    $\begingroup$ Just a helpful note to the one who asked the question: This is the exact same thing that happens when they reach terminal velocity before they release the parachute. $k$ changes because the parachute increases drag, that is why this is all induced again. $\endgroup$ – JMac Feb 11 '17 at 15:15
  • $\begingroup$ @Steeven, thanks for taking the time to edit. It looks much better now. I am guilty that I am still not used to the symbolic formatting techniques and did not take the time to write it out carefully. $\endgroup$ – Abhijeet Melkani Feb 11 '17 at 15:28
  • $\begingroup$ In addition to @JMac's comment: When the parachute opens, it can be considered as a change in $k$ - or it can be considered as just another, extra drag force with $k_{parachute}v$ that must be subtracted as well, $$ma=mg-k_{person}v-k_{parachute}v$$ $\endgroup$ – Steeven Feb 11 '17 at 15:30
  • $\begingroup$ @A.Melkani You are welcome, and no worries. If you are interested, click the edit button and see how \$...\$ symbols make math formatting. $\endgroup$ – Steeven Feb 11 '17 at 15:31
  • 1
    $\begingroup$ I think I'll just add a line in the post to look at the comments. It is easy to enough to understand from the comments itself IMO. $\endgroup$ – Abhijeet Melkani Feb 11 '17 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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