# How does the air resistance affects the graph of a free falling body?

Dec 2, 2017

See explanation

#### Explanation:

The differential equation of the free fall with air resistance is :

$m \frac{\mathrm{dv}}{\mathrm{dt}} = m g - c {v}^{2}$

with

m = mass of falling object
g = gravity constant = 10 m/s² (9.81 but we take 10 for simplicity)
c = constant depending on the Cx value of the object among others
v = vertical velocity downwards of the falling object in m/s

The solution is

h = (-m/(2c)) ln(1-a²)
with
$a = \frac{b - 1}{b + 1}$
and
$b = \exp \left(2 \frac{t}{\sqrt{\frac{m}{c g}}}\right)$

If we take

$\frac{2}{\sqrt{\frac{m}{c g}}} = 1$

we get

$h = - 20 \ln \left(1 - {\left(\frac{{e}^{t} - 1}{{e}^{t} + 1}\right)}^{2}\right)$

with h the height fallen, measured downwards.
We have to compare this with the frictionless fall :

$h = 5 {t}^{2}$

Clearly the latter is a parabole, while the former is flatter.
For the rest there is not much to see about the graphics, but i have added a graph anyway (the red line is frictionless fall, the yellow with air resistance) :