Solving Skydiving Problems: Acceleration & Air Resistance

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Homework Help Overview

The discussion revolves around a physics problem related to skydiving, specifically focusing on the concepts of acceleration and air resistance. The original poster seeks assistance in determining the acceleration of sky divers when the upward force of air resistance is a fraction of their weight, as well as the force of air resistance when descending at constant speed after deploying a parachute.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between net force, weight, and acceleration, questioning how to set up the equations correctly. There is discussion about the implications of constant speed on acceleration and net force.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on understanding net force and its relation to acceleration. There is an ongoing exploration of the correct application of physics principles, and some participants are clarifying their understanding of the problem setup.

Contextual Notes

There is a focus on ensuring clarity in the definitions and calculations involved, with participants questioning assumptions about directionality in their calculations. The original poster is navigating through the problem without having all the necessary information clearly defined.

confusedaboutphysics
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(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?
________ m/s2 (downward)

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4-34.
_____ N (upward)

i need help on starting this problem...like what equation(s) would i use??

thanks so much!
 
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confusedaboutphysics said:
like what equation(s) would i use??

Instead of immediately looking for an equation, try thinking about the problem.

(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?

If the sky divers are subjected to an upward force equal to one-fourth of their total weight, then what is the net force acting on them?

Once you determine that, then think about what law of physics relates net force (which you will have determined by this time) to acceleration (which is what the question is asking for).

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4-34.

If the divers' speed is constant, then what is their acceleration? What then can you say about the net force acting on them?
 
can someone tell me if I'm doing this right for part A?

Fnet = 102(9.8) - (.25)(102)(9.8)
Fnet = 999.6 - 249.9
Fnet = 749.7 = ma

so if I'm doing this right..where do i go from here?
 
confusedaboutphysics said:
can someone tell me if I'm doing this right for part A?

Fnet = 102(9.8) - (.25)(102)(9.8)
Fnet = 999.6 - 249.9
Fnet = 749.7 = ma

So far, so good. Just make sure that you understand that you have elected to take the downward direction as positive. Nothing wrong with that, but you'll find that most physics books choose the opposite convention.

so if I'm doing this right..where do i go from here?

What does the question ask you for?
 
so i know I'm trying to find the acceleration..so would i plug in the answer i got for the mass?

Fnet = 749.7 = ma
Fnet = 749.7a?? but how can i solve from here if i don't know the Fnet and I'm looking for a?
 
You do have the net force! You even stated it:

Fnet = 749.7 = ma

So you now know 749.7 = ma or a = 749.7/m

Make sure you understand the definition of net force.

You set 749.7 = m which isn't correct - think of units.
 
Last edited:
thanks tom and cscott! i understand it much better now
 
Not a problem. :smile:
 
confusedaboutphysics said:
Fnet = 749.7 = ma
Fnet = 749.7a??

Have you taken algebra? You can't just move the "a" from one side of the equals sign to the other like that.
 

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