Falling: If resultant force downwards follows the equation F = mg - kv^2, acceleration is a = g - (k/m)v^2

UKEng20

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I've been putting together a two part equation for showing the motion of an object falling before and after deploying a parachute for a personal project but I keep getting stuck at this modelling stage.

If resultant force downwards follows the equation:
F = mg - kv^2,

acceleration is:
a = g - (k/m)v^2

Here, k is the coefficient determining drag force (coefficient of friction of chute* effective area* density of air*1/2).

Any help in obtaining values for x and v using differential equations?

Cheers!
 

MarkFL

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Observe that acceleration is the time rate of change of velocity and then write::

\(\displaystyle \d{v}{t}=g-\frac{k}{m}v^2\) where \(v(0)=v_0\)

Note that the ODE is separable, and integration will likely involve partial fraction decomposition. Can you proceed?
 

UKEng20

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💥 Moderator Note: Broken link removed

Thank you! Does this fraction match the one you'd expect me to be decomposing?
 

MarkFL

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That's virtually unreadable on my end. Are you familiar with \(\LaTeX\)?
 

mmm4444bot

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💡 In the forum guidelines, there's a link to a site which explains how to format math as text. Basically, put grouping symbols around all numerators/denominators which contain more than a single number or symbol.

EG:

(x^2 + 15)/[(x + 3)(x^2 + 3)] = A/(x + 3) + B/(x + 3)^2 + (Cx + D)/(x^2 + 3)

😎
 

UKEng20

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Apologies. The fraction I meant to write was:

1/(g-(k/m)v^2)

Alternately:

(m/k)/(mg/k-v^2)
 
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