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(t_int) 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)t_int + 1/2at_int²

0=t_int(usinσ + 1/2at_int)

t_int=0
or
t_int=-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 t_intl(t I had found from looking at motion in the y(-4.9usinσ) to get the point at which the time ended(t_endend)

t_end=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)

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

then using the the equation d=v/t I put in v_avg for the v, t_int 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=[itex]\sqrt{2kd/(-39.2ksinσcosσ+48.02sinσ)}[/itex]

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

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

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

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

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.