Thanks for your reply, I know completely understand! This is just a question I have been given by my maths teacher. Definitely our high school grade maths.

When it comes to solving $x¨=g-kv^2$ I'm completely lost

This is why I think my solution does not satisfy your requirements. Have you worked with differential equations?

Yes, towards rectilinear motion We have done these also: 1. Integration by parts 2. Variable forces 3. Newton's laws and antiderivatives of such etc 4. Substitution where the derivative is present in the integrand 5. Linear substitution 6. Trigonometric integration 7. Definite integrals Pretty much all the stuff assumed by my teacher to be able to cover this question. Thanks again!

I am not sure how just knowledge of integration by parts will help you here. I will keep thinking. Do you have any idea how the problem should be solved? initial equations? - failing that I will give a short edit on solving the equation you start in the comment above.

Sorry i was messing up my edits consistently

NP. Are you happy with the chain rule $\dfrac{dv}{dt} = \dfrac{dx}{dt}\times \dfrac{dv}{dx}$

Yes i understand the chain rule 110%

I've picked up on some implicit differentiation in some of the derivation above also

Gotcha

Ok, so you understand the chain rule so from that you can integrate the ode.

integrate the ode?

I am running out of ideas how to prove the solution. So essentially what I do is state $$\dfrac{d^2 x}{dt^2} = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = v\frac{dv}{dx} = \frac{1}{2}\dfrac{d}{dx}v^2$$

then you can define $y = v^2$ then you have a linear ode $\frac{1}{2}\frac{dy}{dx} = g - ky$ which you may be able to solve?

i understand all of that besides where you got the $\frac{1}{2}\dfrac{d}{dx}v^2$

$\frac{d}{dx}v^2 = 2v\frac{dv}{dx}$ this is implicit differentiation. But we require only $v\frac{dv}{dx}$ so to achieve this we divide both sides by 2.

This leads to $v\frac{dv}{dx} = \frac{1}{2}\dfrac{d}{dx}v^2$

oh i see now, then we rearrange to make dx the subject such that it equals $$\frac{dv}{g-kv^2}$$

then you could replace dx in the integrand and then integrate using the value of dx

almost remember that you have $\frac{1}{2}\dfrac{d}{dx}v^2$

but essentially you are correct. Then you move the $dx$ over to the rhs

i see

the next thing i don't understand is the re-arrangement

i've figured so far that since $F=ma$, F=\frac{1}{2}\dfrac{d}{dx}v^2 = mg-kv^2$

since $m=1$, $mg=9.8$

therefore, $9.8-kv^2 = \frac{1}{2}\dfrac{d}{dx}v^2$

ok so we will keep it as mg or just g

A real mathematician (which I am not) would disagree with my terminology - but you divide both sides by the right hand side leaving unity on the rhs

so that we can put this all into perspective are you able to edit your original post? I will than be able to see the process better

You are correct in terms of the last equation

A real mathematician (which I am not) would disagree with my terminology - but you divide both sides by the right hand side leaving unity on the rhs

So in terms of integrating what do i do next after we have this relationship?

You can relabel terms to make it clearer .. So substitute $y = v^2$ and see what you get

oky doke

AHA! I have it!

Thankyou very very much!

However, i'm having trouble with where 2x+C has come from, i haven't done an integration like that before

It came from when I moved the half from the one side to the right.

The C is an integration constant

When solving integrals you have definite and indefinite? Basically with limits and without. Since I did an integration without limits I needed a constant (which you find by inserting in the starting information)