Skydiving: Falling 1.2 km at 9.8 m/s^2

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Homework Help Overview

The discussion centers around a skydiving scenario where a person falls from a height of 1.2 kilometers, with participants exploring the physics of free fall, including gravitational acceleration and velocity calculations. The subject area involves kinematics and the effects of gravity, as well as considerations of air resistance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational acceleration and velocity, with some suggesting the use of conservation of energy to find velocity. Others express confusion about how to calculate velocity without time and question the height at which parachutes are deployed.

Discussion Status

The discussion is ongoing, with various approaches being explored, including kinematic equations and conservation of energy. Some participants have raised concerns about the neglect of air resistance in the calculations. There is no explicit consensus on the best method to solve the problem, and multiple interpretations are being considered.

Contextual Notes

Some participants note that the question may be part of an introductory physics course, which could influence the level of complexity expected in the responses. There is also mention of a template that was not followed, which may have affected the clarity of the original question.

ADH
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A person goes skydiving. The plane is one kilometer up and they decide to fall a height of 1.2 kilometeres before opening their chutes. How fast are they going at that point?

Isn't it 9.8 m/s^2 because gravity has to be dragging them down at that rate?

Thank you!
 
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ADH said:
Isn't it 9.8 m/s^2 because gravity has to be dragging them down at that rate?

They will be accelerating at that rate, but that will not be how fast they are going. To get how fast, you need to find the velocity at that point.

Conservation of energy will solve this quickly :wink:
 


So velocity is distance / time. I have the distance, but don't have the time. How am I supposed to solve for velocity then?
 


Um. Why do they open their parachutes after falling 1.2km when they are only 1.0km high?
 


Phrak said:
Um. Why do they open their parachutes after falling 1.2km when they are only 1.0km high?

I meant 1/2. That's what the title says. I made a mistake in the description.
 


99 percent of terminal velocity, with prone attitude, for a typical body type takes about 550 metres. 90% of terminal velocity happens after falling about 375 metres.
 


ADH said:
So velocity is distance / time. I have the distance, but don't have the time. How am I supposed to solve for velocity then?

Consider the energy before and the energy after.

Conservation of energy: KE+PE = constant.


As for the why the parachute is opened, I am not sure about the height they have to open their chutes at, but if they open it without reaching terminal velocity, then bad things happen.
 


Why are you ignoring air friction, rock.freak. Skydivers don't fall in a vacuum.
 


Phrak said:
Why are you ignoring air friction, rock.freak. Skydivers don't fall in a vacuum.

This is probably a few weeks into an intro physics course, chill out. You got to walk before you can run.
 
  • #10


Phrak said:
Why are you ignoring air friction, rock.freak. Skydivers don't fall in a vacuum.

For two reasons, one what Pengwuino said and two, had air resistance been included, more information would need to be given I believe.

Pengwuino said:
This is probably a few weeks into an intro physics course, chill out. You got to walk before you can run.
 
  • #11


I believe it is like this:

Vi=0m/s a=9.8m/s^2 d=500m Vf=?

So we use this formula:

Vf^2= Vi^2 + 2ad

Once you plug in the numbers and calculate, it should come out to 99 m/s

However, I just started learning Physics yesterday, so do not take my word for it.
 
  • #12


Thanks OThePestO.

But really, this is a Physics forum. How is it that nobody knows the answer? This is a basic physics question, it doesn't involve extremely complicated calculation yet.
 
  • #13


Its a simple kinematic problem: don't listen to the others about conservation of energy and drag, because you have not learned that yet.

y = (vi)t - (1/2)gt^2

since vi = 0; your jumping out of a plane you get

y = -(1/2)gt^2set y to 500m (1/2 km) and solve for the time. Once you have time you can plug it in a kinematic equation to find the velocity.

Velocity is the derivative of position (assuming your class is calculus based)

v = dy/dt = -(gt)

plug n solve
 
  • #14


ADH said:
Thanks OThePestO.

But really, this is a Physics forum. How is it that nobody knows the answer? This is a basic physics question, it doesn't involve extremely complicated calculation yet.

Do you have an answer to reference with? If no one else provides a better solution, I would just go with mine, assuming this is for marks. What are the current formulas you are working with, as I assume you would be required to show how you came to use the formula I provided if you had not yet learned it.
 
  • #15


ADH said:
Thanks OThePestO.

But really, this is a Physics forum. How is it that nobody knows the answer?

We are forbidden to give you the answer. We can help you arrive at the correct answer.


Also, there was a template you were given in which you were supposed to supply the question. You deleted it.
 
  • #16


rock.freak667 said:
For two reasons, one what Pengwuino said and two, had air resistance been included, more information would need to be given I believe.

Yeah, but I can't agree. If professors will say what they mean and mean what they say...

Or maybe you're wrong and this is all about terminal velocity for the bright students. Let me consult my Ouji board...
 
Last edited:
  • #17


Phrak said:
Yeah, but I can't agree. If professors will say what they mean and mean what they say...

Or maybe you're wrong and this is all about terminal velocity for the bright students. Let me consult my Ouji board...

Not enough information is given in the question to utilize finding the terminal velocity.
 
  • #18


DaveC426913 said:
We are forbidden to give you the answer. We can help you arrive at the correct answer.


Also, there was a template you were given in which you were supposed to supply the question. You deleted it.

The template was terrible. It was much easier to just write my question. You're not allowed to give answers on this forum? Is that a joke? If I've been working hard on a problem, and all of the help isn't helping, it would help much more to give me the answer so I would have something to work towards.
 
  • #19


ADH said:
The template was terrible. It was much easier to just write my question. You're not allowed to give answers on this forum? Is that a joke? If I've been working hard on a problem, and all of the help isn't helping, it would help much more to give me the answer so I would have something to work towards.

Thus far, I see that three replies were given that would attempt to aid you, but since they all have differing things which depend on what you've learned, it is hard to say.

They were
Conservation of Energy
Using kinematics
Terminal velocity

Which one have you learned in your class?


Also, you didn't really post what you attempted, so we don't really know along what lines you tried which is why everyone has a differing input.
 
  • #20


Look at my post, follow the steps and you will arrive at the correct answers.
 
  • #21


Pengwuino said:
This is probably a few weeks into an intro physics course, chill out. You got to walk before you can run.

why would they have terminal velocity either. That is at least mechanics 1 and this seems like grade 10 physics
 

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