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

Please show 2 separate graphs in answering your question.

1 Answer
Dec 2, 2017

Answer:

See explanation

Explanation:

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

#m {dv}/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 = (b - 1)/(b + 1)#
and
#b = exp(2t/sqrt(m/(cg)))#

If we take

#2/sqrt(m/(cg)) = 1#

we get

#h = -20 ln(1-((e^t-1)/(e^t+1))^2)#

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) :

enter image source here