Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

So I have these two parachute problems that have been giving me problems any aid you can provide would be much apreciated!

1. The problem statement, all variables and given/known data

A)A probe is to land on the surface of mars and I am required to find out the diameter of the parachute.

Given constance:

m=40kg

g=3.75m/s^{2}

v_{0}=100m/s

v_{f}= 3m/s

x_{0}=100m

Pair=2/3*1.1774

Cd=1.4

A=pi*D^{2}/4 (where D is the unknown diameter of the parachute)

2. Relevant equations

m[tex]\frac{dV}{dt}[/tex]=mg-[tex]\frac{1}{2}[/tex]Pair*Cd*AV^{2}

3. The attempt at a solution

I've seprated the variables and simplified to get:

g-[tex]\frac{k}{m}[/tex]*V^{2}*dV=dt

where k=[tex]\frac{1}{2}[/tex]PairCdA

[tex]\sqrt{}[/tex]

So i decomposed the function for integration:

[tex]\int[/tex][tex]\frac{\frac{1}{\sqrt{g}}}{\sqrt{g-\sqrt{\frac{k}{m}}V}}[/tex]+[tex]\frac{\frac{1}{\sqrt{g}}}{\sqrt{g+\sqrt{\frac{k}{m}}V}}[/tex]

(ignore the second square root on top of the K/m that comes from G, it's only supposed to be on g)

and so that I've integrated to get the following:

[tex]\frac{1}{\sqrt{g}*\sqrt{\frac{k}{m}}}(Ln(\sqrt{g}-\sqrt{\frac{k}{m}}V))+Ln(\sqrt{g}-\sqrt{\frac{k}{m}}V))=t+c[/tex]

At this point I understand that I can raise everything to E and cancel the +-[tex]\sqrt{\frac{k}{m}}V[/tex] but from here I am unsure (also not 100% on the integration)

Also, I have a similar problem however it's not falling, rather slowing down from a parachute (rocket car slowing down from parachute), where I need to find the distance required to stop the car.

Where I was given m[tex]\frac{dV}{dt}[/tex]=-[tex]\frac{1}{2}[/tex]Pair*Cd*AV^{2}

So I integrated as a seperable (don't think this is the right way though) to get

[tex]\frac{-1}{kV}=t+c[/tex] (a check or direction for this one is all I need)

Thank you!

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# Homework Help: Two Parachute Problems

Can you offer guidance or do you also need help?

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