The velocity of a skydiver falling to the ground

In summary, the velocity of a skydiver falling to the ground is governed by equations that take into account the skydiver's mass, the acceleration due to gravity, and the drag coefficient.
  • #1
Jamin2112
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The velocity of a skydiver falling to the ground ...

Homework Statement



SORRY FOR SO MANY QUESTIONS! I'VE BEEN WORKING THROUGH EACH OF THESE UNTIL I'M COMPLETELY STUCK, SO DON'T THINK I'M JUST TRYING TO SUCK ANSWERS OUT OF YOU FOLKS!

The velocity v(t) of a skydiver falling to the ground is governed by

m(dv/dt)=mg-kv2

where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag coefficient, and v(t)≥0

Homework Equations



Going to go separable on this one

The Attempt at a Solution



Problem #1: Something seems to be wrong. Since v(t)≥0, the problem is orienting positive velocity in the downwards direction. Wouldn't the drag coefficient, k, therefore be less than zero? It's pushing against the skydiver. Seems like common sense to me.

Problem #2: I've always been told that mass has nothing to do with the speed at which an object falls. Why is it in an equation for change in velocity, dv/dt?

m(dv/dt)=mg-kv2

==> dv/dt=g-(k/m)v2
==> dv/dt =(-k/m)(-gm/k + v2)
==> dv/(-gm/k + v2)=(-k/m)dt

Problem #3: Now it looks like I could use dx/(a2+x2)=(1/a)tan-1(x/a) ... but then I'd have a= +/- √(-gm/k), which would be an imaginary number since g,m, and k are all positive!
 
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  • #2


prob #1, the negative in the equation takes care of that & makes k>0

prob #2, if you don't include the drag the m will cancel out, however drag force is generally proportional to surface area whilst acceleration is proportinal to mass, so you can no longer cancel the m.

To see that consider the terminal velocity of a ball bearing & feather with the same mass. The surface area of the feather is much greater, so the drag force is greater and the feather will have a much lower terminal velocity.
 
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  • #3


adding to #1 the best way to check is to draw a Free body diagram with all the forces
 
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  • #4


for #3 how about trying the intergal
[tex] \int \frac{v.dv}{1-v^2} [/tex]
and using the subtitution
[tex] u = 1-v^2 [/tex]
then
[tex] du = -2v.dv [/tex]
and the integral becomes
[tex] \int \frac{-du}{2u} [/tex]

I haven't followed this through but think it looks promising, though you'll have to add constants and stuff
 

1. What factors affect the velocity of a skydiver falling to the ground?

The velocity of a skydiver falling to the ground is affected by several factors, including the force of gravity, the mass of the skydiver, and the air resistance.

2. How does the velocity of a skydiver change as they fall towards the ground?

As a skydiver falls towards the ground, their velocity increases due to the force of gravity. However, as they approach terminal velocity, their velocity will remain constant due to the balancing forces of gravity and air resistance.

3. What is terminal velocity and how does it relate to the velocity of a skydiver?

Terminal velocity is the maximum speed that an object, such as a skydiver, can reach when falling through a fluid, such as air. It is the point at which the force of air resistance is equal to the force of gravity, resulting in a constant velocity.

4. Can a skydiver increase their velocity while falling?

Yes, a skydiver can increase their velocity while falling by changing their body position to reduce air resistance. For example, by diving downwards, they can increase their velocity, while spreading their arms and legs can increase air resistance and slow them down.

5. How does altitude affect the velocity of a skydiver?

The higher the altitude, the longer a skydiver has to accelerate due to the greater distance to the ground. Therefore, the velocity of a skydiver will be higher at higher altitudes compared to lower altitudes.

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