Projectile trajectory

[Function]

Hi guys

 

Everyone knows (or should know,) this pretty important formula:

 

[math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2\cdot v_0^2\cdot\cos^2{\theta}}[/math]

 

Now, I don't know if I'm physically allowed to do this, but I derived that, keeping in mind that every unit, independent of time, is constant:

 

[math]\frac{dy}{dt}=\tan{\theta}\cdot \frac{dx}{dt}-\frac{2\cdot g}{2\cdot v_0^2\cdot\cos^2{\theta}}\cdot\frac{dx}{dt}[/math]

 

[math]v_y=v_x\cdot\left(\tan{\theta}-\frac{g}{v_0^2\cdot\cos^2{\theta}}\right)[/math]

 

Now,

 

[math]v_{y,0}=v_0\cdot\sin{\theta}[/math]

 

and

 

[math]v_{x,0}=v_0\cdot\cos{\theta}[/math]

 

Am I thus allowed (?) to say that, using the formulas of kinematics:

 

[math]v_y=v_{y,0}-g\cdot t = v_0\cdot\sin{\theta}-g\cdot t[/math]

 

[math]v_x=v_{x,0}=v_0\cdot\cos{\theta}[/math]

 

Resulting in:

 

[math]v_0\cdot\sin{\theta}-g\cdot t = v_0\cdot\cos{\theta}\cdot\left(\tan{\theta}-\frac{g}{v_0^2\cdot\cos^2{\theta}}\right)[/math]

 

[math]v_0\cdot\sin{\theta}-g\cdot t = v_0\cdot\sin{\theta}-\frac{g}{v_0\cdot\cos{\theta}}[/math]

 

[math]t=\frac{1}{v_0 \cdot \cos{\theta}}[/math]

 

Now, if everything I did was 'legal' and correct, what [math]t[/math] is this? Is it the time the projectile needs to complete its trajectory? Or is just about everything I did wrong?

 

Thanks!

 

Function

Edited August 19, 2014 by Function

[studiot]

It would help if you got your Latex errors mended.

 

However is angle theta not the initial angle of projection?

 

So are you expecting this to remain constant throughout the trajectory?

[Function]

Errors fixed.

Oh God how did I oversee that... But what's [math]\dfrac{d\theta}{dt}[/math]? Is it [math]\omega[/math]?

And following this reasoning, [math]v_0[/math] should also be derived?

 

Edit: reread your reaction; theta is the initial angle; should it be derived?

 

Edit (bis): numerous mistakes in deriving. Shall review.

Edited August 19, 2014 by Function

[swansont]

It would help if you got your Latex errors mended.

 

However is angle theta not the initial angle of projection?

 

So are you expecting this to remain constant throughout the trajectory?

 

The initial angle is a constant for the problem. It's the initial angle

Hi guys

 

Everyone knows (or should know,) this pretty important formula:

 

[math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2\cdot v_0^2\cdot\cos^2{\theta}}[/math]

 

Now, I don't know if I'm physically allowed to do this, but I derived that, keeping in mind that every unit, independent of time, is constant:

 

[math]\frac{dy}{dt}=\tan{\theta}\cdot \frac{dx}{dt}-\frac{2\cdot g}{2\cdot v_0^2\cdot\cos^2{\theta}}\cdot\frac{dx}{dt}[/math]

 

 

…

 

Now, if everything I did was 'legal' and correct

 

It's not. You took the derivative of x² and the result should be 2x dx/dt

 

Your final result has a time equal to an inverse of a speed. The units don't match up, so it is necessarily incorrect.

[Function]

After new derivation, I get:

 

[math]t=\frac{x}{v_0\cdot\cos{\theta}}[/math]

 

Well... That's pretty useless.. (Units match up, though )

Edited August 19, 2014 by Function

[swansont]

Hi guys

 

Everyone knows (or should know,) this pretty important formula:

 

[math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2\cdot v_0^2\cdot\cos^2{\theta}}[/math]

 

 

One nit: I don't know this formula. i.e. I don't have it committed to memory, and I'm a physics PhD. I can easily derive it, though, which IMO is a lot more important and powerful.

After new derivation, I get:

 

[math]t=\frac{x}{v_0\cdot\cos{\theta}}[/math]

 

Well... That's pretty useless.. (Units match up, though )

 

Not at all. It tells you how long it takes to go a distance x given an initial launch speed and angle.

[studiot]

 

The initial angle is a constant for the problem. It's the initial angle

 

 

It is also the slope of the X-Y trajectory at t=0, that is it is

 

[math]{\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}}[/math]

 

 

The point I wanted to establish was that this slope changes throughout the trajectory.

 

function was there anything else you wish to discuss about this subject?

 

You can derive further information from your formula if you wish.

Edited August 19, 2014 by studiot

[Function]

 

It is also the slope of the X-Y trajectory at t=0, that is it is

 

[math]{\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}}[/math]

 

 

The point I wanted to establish was that this slope changes throughout the trajectory.

 

function was there anything else you wish to discuss about this subject?

 

You can derive further information from your formula if you wish.

 

I guess not, atm. It was the consequence of a strange equation I got by a wrong derivation... Thanks.

[imatfaal]

Whilst "shutting up and calculating" might be the way for quantum mechanics (I don't really know is this is the case or as I expect a massive exaggeration) in classical mechanics it really is a mistake not to understand what each line of your working means and implies. If I have read it correctly you quickly had an expression that showed that vertical component of velocity varied only with the horizontal component of velocity (which doesn't vary with time) - but in a trajectory that is clearly not the case; it goes up and it comes back down - how can the y velocity not vary with time. Checking that you have maintained a physically sensible state of affairs allows you to catch errors in maths - and vice versa of course.