# Air Resistance solution (is it valid)?

Last semester I did a project for grade 12 physics where I had to account for air resistance in projectile motion. I just wanted to post my solution here and as I never found out the merit of the solution, my teacher said that it came out to around the right anser but he said he didn't understand exactly what I did.

In this post I am trying to find initial velocity given distance, the angle of departure, air density, the drag coefficient, the reference area, and the mass.

The first assumption I made is that I would launch my projectile at zero height.

My second assumption was that since the trajectory would be relatively symmetrical* the force of drag would exert the same magnitude of force on the way up as on the way down, and therefore the time that the object spends in the air would be the same amount of time that the same object would spend if there was no air resistance.

*I'm skeptical about how accurate this is?

My third assumption was that their would be turbulent flow around the projectile

My fourth assumption is that air density, coefficient of drag and Reference don't change significantly while the projectile is in the air.

My fifth assumption was that their was no wind

Using this assumption I could then find the time(tint) in terms of the intial velocity(u), and the angle of departure(σ) and that the object was in flight using the formula.
d=0
a=9.8
u=?

d= (usinσ)tint + 1/2atint2

0=tint(usinσ + 1/2atint)

tint=0
or
tint=-4.9usinσ

I then went on to do air resistance in the horizontal.

F=1/2pv2CA

and rearanged to

a=(pv2CA)/(2m)

I then assumed that density of air(p), the drag coefficient(C), the reference area (A) and the mass would all be constant during the time of flight. Therefore I got some constant(k) times velocity squared is equal to the acceleration at any given point in flight.

k=pCA/2m
a=kv2

the derivative equation of this would result in a graph that measured time as a function of velocity because the slope on the graph with acceleration in the y and velocity in the x is time.

Therefore

T(v)= 2kv

Then I put in my initial velocity in the horizontal (ucosσ) to get the point at which my time started.

T(ucosσ) =2kucosσ

I then added the time interval tintl(t I had found from looking at motion in the y(-4.9usinσ) to get the point at which the time ended(tendend)

tend=2kucosσ-4.9usinσ

I then found the velocity at this point(v2

v2=(2kucosσ-4.9usinσ)/2k

I then found the average velocity by adding u and v2 together and dividing by two(since velocity is measured in the x the change is constant over the interval and the average can be found in this way)

vavg=(((2kucosσ-4.9usinσ)/2k)+ucosσ)/2

then using the the equation d=v/t I put in vavg for the v, tint for time, and my distance in the horizontal direction for d.

d=(((2kucosσ-4.9usinσ)/2k)+ucosσ)/2)/-4.9usinσ

d=(-4.9usinσ*(2kucosσ-4.9usinσ)/2k)+2kucosσ/2k)/2

d=(-9.8ku2sinσcosσ+24.01u2sin2σ--9.8ku2sinσcosσ/2k)/2

d=(2u2(-19.6ksinσcosσ+24.01sinσ))/2k

u=$\sqrt{2kd/(-39.2ksinσcosσ+48.02sinσ)}$

or
since
k=pCA/2m
and
39.2=4g
48.02=2(g/2)2

u=$\sqrt{\frac{dpCA/m}{sinσ*(-4gpCAcosσ/2m+2(g/2)^{2})}}$

u=$\sqrt{\frac{dpCA}{msinσ*(-4gpCAcosσ/2m+4m(g/2)^{2}/2m)}}$

u=$\sqrt{\frac{dmpCA}{sinσ*(-4(gpCAcosσ-m(g/2)^{2}))}}$

Sorry for any stupid mistakes I lost my solution and doing this on the computer screws me up, plus I am prone to make them. Also sorry for any equations that I could have typed better I did not figure out that the sigma symbol gives tools for complex equations.

should this post be in the homework questions section?

Last edited:

## Answers and Replies

mfb
Mentor
My second assumption was that since the trajectory would be relatively symmetrical* the force of drag would exert the same magnitude of force on the way up as on the way down, and therefore the time that the object spends in the air would be the same amount of time that the same object would spend if there was no air resistance.
For high speeds and small σ, the approximation might be fine. In general, it can be problematic - air drag reduces time of ascent (drag+gravity in the direction), and increases time of descent (drag+gravity in opposite direction).

the derivative equation of this would result in a graph that measured time as a function of velocity because the slope on the graph with acceleration in the y and velocity in the x is time.
I don't understand that.

With a=kv^2 and the approximation that the velocity does not change too much during the flight, you can use v0 = vend+kvend2*t as a first approximation. A better approximation is given by the solution of
v0 = vend+k(vend2+v02)*t
And if you need more precision, you should give up the second assumption.

I'm trying to understand why higher speeds and small σ make the approximation more fine. At smaller σ won't their be more drag in the horizontal and thus distort the trajectory more than a higher σ. At σ=90 shouldn't this approximation be the most accurate since the force exerted on the way up be the same as on the way down and thus the time reduced on the ascent would cancel with the time gained on the way down. Also why does speed increase the accuracy of the approximation. Wouldn't the error be proportional.

slope=$\frac{Δy}{Δx}$

slope=$\frac{Δa}{Δv}$

slope=$\frac{m/s^{2}}{m/s}$

slope=$\frac{m}{s^{2}}*\frac{s}{m}$

slope=$s$

slope=$t$

Thus since the 1rst derivative plots slope as a function of the original independent, the derivative would plot time as a function of velocity.

With a=kv^2 and the approximation that the velocity does not change too much during the flight, you can use v0 = vend+kvend2*t as a first approximation. A better approximation is given by the solution of
v0 = vend+k(vend2+v02)*t
And if you need more precision, you should give up the second assumption.

Assuming that the velocity in the horizontal direction doesn't change much, conflict with the approximation that the time of flight is the same as if their was no air resistance since you said that that approximation might be fine under high velocities which would produce a greater velocity change.

mfb
Mentor
Hmm, higher speed was a bit unprecise. Smaller influence of air drag would be better.
Smaller σ lead to shorter flight times and therefore a smaller influence of air drag, too. A vertical flight completely ruins your approximations, as you get all sorts of higher-order effects. In addition, the velocity changes much during flight.

Thus since the 1rst derivative plots slope as a function of the original independent, the derivative would plot time as a function of velocity.
I have no idea what you mean here. Derivative of what with respect to what, and plot of what?

The time of flight depends on the vertical velocity - if that is small, the relative velocity change in the horizontal direction might be small (but depends on your parameters).