(Vf)^2 = (Vi)^2 + 2a Δs
Proof

Question

solomonzelman · Answer

Attached below.

solomonzelman · Answer

$v_{\rm f}^{~2}=v_{\rm f}^{~2}+2{\rm a}\Delta {\rm s}$  proof is attached above.

solomonzelman · Answer

I might go ahead and post some other things later... 
This is it for now. ((let's keep the thread clean, unless I have made some mistakes or said something not precisely correct))

michele_laino · Answer

good work! :)

solomonzelman · Answer

Yes, nice sketch. Visualization is helpful, and in physics always possible, unlike math:

farcher · Answer

A good derivation but I like to draw a sketch graph and the follow what is essentially your derivation as shown below:

|dw:1451927108532:dw|

astrophysics · Answer

Good stuff, looking forward for more! @SolomonZelman 

Haha every time I think about these equations I first think about deriving with the graphs as you've presented @Farcher

solomonzelman · Answer

Sure, you can recall that acceleration is the second derivative of position... there is a route from there.

solomonzelman · Answer

Posting directly is so much easier... I am so used to working with latex, and programs that use latex are expensive. I always have to adjust myself to equation editor in word and make good ooking combinations of fonts and sizes.... I will type them up directly into the thread.

solomonzelman · Answer

$\color{#f000f0 }{ \displaystyle \rm Projectile~Motion~Path ~~(proof) }$

NOTATIONS:
$\color{#000000 }{ \displaystyle v_{i}  }$	$-$	initial speed ((scalar))
$\color{#000000 }{ \displaystyle \theta_{i}  }$	$-$	initial angle of the trajectory
$\color{#000000 }{ \displaystyle v_{ix}  }$	$-$	initial velocity in the horizontal
$\color{#000000 }{ \displaystyle v_{iy}  }$	$-$	initial velocity in the vertical
$\color{#000000 }{ \displaystyle g  }$	$-$	gravity
$\color{#000000 }{ \displaystyle t  }$	$-$	time (kind of obvious:) )
$\color{#000000 }{ \displaystyle x_{0}  }$	$-$	initial x-value (here - initial horizontal position)
$\color{#000000 }{ \displaystyle y_{0}  }$	$-$	initial y-value (here - initial vertical position)

$\color{#000000 }{ \displaystyle v_{x}(t)  }$	velocity in the horizontal
		(as a function of time)$\tiny \[0.7em]$
$\color{#000000 }{ \displaystyle v_{y}(t)  }$		velocity in the vertical
		(as a function of time)$\tiny \[0.7em]$
$\color{#000000 }{ \displaystyle x(t)  }$		horizontal displacement
		(as a function of time)$\tiny \[0.7em]$
$\color{#000000 }{ \displaystyle y(t)  }$		vertical displacement
		(as a function of time)$\tiny \[1.5em]$
$\color{#0000ff }{ \displaystyle y(x)  }$		vertical displacement (function of x)
		(will be used last for the path-function)

PRESUMED FORMULAS:
$\color{#000000 }{ \displaystyle v_{ix}=v_{i}\cos\theta_{i}   }$
$\color{#000000 }{ \displaystyle v_{iy}=v_{i}\sin\theta_{i}   }$

$\color{#000000 }{ \displaystyle v_x(t)=v_{ix}  }$		(horiz. velocity doesn't change)
$\color{#000000 }{ \displaystyle v_y(t)=v_{iy} -gt }$	(vert. velocity is affected by gravity)

PROOF:
Vertical velocity,
$\color{#000000 }{ \displaystyle v_y(t)=v_{iy} -gt }$
After integrating both sides with respect to time (t), you get,
((vertical velocity, when integrated, is vertical displacement))

$\color{#000000 }{ \displaystyle y(t)+C=v_{iy}t -\frac{1}{2}gt^2+C }$
$\color{#000000 }{ \displaystyle y(t)=v_{iy}t -\frac{1}{2}gt^2+C }$

The +C, is the initial vertical displacmenent/shift that the particle may have before being fired:

$\color{#000000 }{ \displaystyle y(t)=v_{iy}t -\frac{1}{2}gt^2+y_0 }$

Let's write everything in terms of initial angle and the speed.
$\color{#000000 }{ \displaystyle y(t)=v_{i}\sin(\theta_i)\cdot t  -\frac{1}{2}gt^2+y_0 }$

We want to model a y=f(x) for the pattern, so, we need to dermine the substitute for x. Recall our "presumed" formula,
$\color{#000000 }{ \displaystyle v_{x}(t)=v_{ix}  }$
We can integrate both sides with respect to t,
(the horizontal velocity, when integrated, is horizontal displacement)

$\color{#000000 }{ \displaystyle x(t)+C=v_{ix} t+C }$
$\color{#000000 }{ \displaystyle x(t)=v_{ix} t+C }$

And in this context, the C is the initial position in the x,
$\color{#000000 }{ \displaystyle x(t)=v_{ix} t+x_0 }$

Usually, we assume that the x-coordinate of the trajectory is 0. THis is how we do it, and this is most convinient. So, keeping in my that $\color{#000000 }{ \displaystyle x_0 =0}$, we can write,

$\color{#000000 }{ \displaystyle x(t)=v_{ix} t }$

(Note that I kept $\color{#000000 }{ \displaystyle y_0 }$, because sometimes projectile problems start above the ground)

And this is what we may substitute for t,

$\color{#000000 }{ \displaystyle \frac{x(t)}{v_{ix} }=t }$

We had,

$\color{#000000 }{ \displaystyle y(t)=v_{i}\sin(\theta_i)\cdot t  -\frac{1}{2}gt^2+y_0 }$

And now this becomes,

$\color{#000000 }{ \displaystyle y(x)=v_{i}\sin(\theta_i)\cdot \frac{x(t)}{v_{ix}}  -\frac{1}{2}g\left(\frac{x(t)}{v_{ix}}\right)^2+y_0 }$


let's re-write the initial horiz. velocity,

$\color{#000000 }{ \displaystyle y(x)=v_{i}\sin(\theta_i)\cdot \frac{x(t)}{v_{i}\cos(\theta_i)}  -\frac{1}{2}g\left(\frac{x(t)}{v_{i}\cos(\theta_i)}\right)^2+y_0 }$

Ok, let's keep in mind that x is a function of time, but will ease on the notation,

$\color{#000000 }{ \displaystyle y(x)=v_{i}\sin(\theta_i)\cdot \frac{x}{v_{i}\cos(\theta_i)}  -\frac{1}{2}g\left(\frac{x}{v_{i}\cos(\theta_i)}\right)^2+y_0 }$

Now, will apply some algebra, and we get,

$\color{#000000 }{ \displaystyle y(x)=x\tan(\theta_i) -\frac{g}{2v_{i}^{~2}\cos^2(\theta_i)}x^2 +y_0 }$

solomonzelman · Answer

Well, if you apply the definitions, then

$\color{#000000 }{ \displaystyle y(x) =\Delta y }$
$\color{#000000 }{ \displaystyle x =\Delta x }$

would be equivalent substitutions
(but, then we would just erase $\color{#000000 }{ \displaystyle y_0 }$ from the equation)

solomonzelman · Answer

$\color{#000000 }{ \displaystyle v_{y}(t)=v_{iy}-gt }$

At the top, the vert. velocity is zero, so let's solve for the time when vert. velocity is zero, to determine at what time the top of the trajectory is reached.
$\color{#000000 }{ \displaystyle 0=v_{iy}-gt }$
$\color{#000000 }{ \displaystyle gt=v_{iy}}$
$\color{#000000 }{ \displaystyle t=v_{iy}/g}$

So, it takes $\color{#000000 }{ \displaystyle v_{iy}/g}$ to reach the top.
((Or, you can rewrite that as $\color{#000000 }{ \displaystyle t=v_{i}\sin(\theta_i)/g}$   ))

solomonzelman · Answer

Since, the trajectory is summetric (this is alwas the assumption made by projectile motion problems), therefore,

$\color{#000000 }{ \displaystyle t_{\rm ~flight}= 2\times t_{\rm ~top}=2\times v_{iy}/g=2\times v_{i}\sin(\theta_i)/g}$

((Since, it takes the same time to go from top to the end as it takes from start to top))

solomonzelman · Answer

I closed the question, but here is some more....

solomonzelman · Answer

This relates to projectile motion range.

REMINDER:
Projectile path is symmetric about the vertex, and vertex is the maximum.

So, I know that
$\color{#000000 }{ \displaystyle v_{y}(t)=v_{iy} -gt}$
(velocity in the y-direction)

To find an expression for vertical displacement/position at time t, I integrate both sides,

$\color{#000000 }{ \displaystyle \int v_{y}(t)~dt=\int~v_{iy} -gt~dt}$

$\color{#000000 }{ \displaystyle  y(t)=v_{iy}t -\frac{1}{2}gt^2+y_0}$

(In this context, +C is the initial vertical position from where the particle starts to travel)
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Then, I know that the time to reach the top
(recall that vertical velocity at the top is zero),

$\color{#000000 }{ \displaystyle 0=v_{iy} -gt}$
$\color{#000000 }{ \displaystyle gt=v_{iy} }$
$\color{#000000 }{ \displaystyle t=v_{iy}/g }$  (time to get to the top)

Or, in terms of the initial angle and initial speed,

$\color{#000000 }{ \displaystyle  y(t)=v_{i}\sin(\theta)t -\frac{1}{2}gt^2+y_0}$
(the vertical position as a function of time)

$\color{#000000 }{ \displaystyle t_{\rm ~top}=v_{i}\sin(\theta)/g }$
(time to reach the top)

So the maximum y, (the range) is,

$\color{#000000 }{ \displaystyle   {\rm Range}=y(v_{i}\sin(\theta))}$

And for that I get,
$\color{#000000 }{ \displaystyle  {\rm Range}=v_{i}\sin(\theta)\left(\frac{v_{i}\sin(\theta)}{g}\right) -\frac{1}{2}g\left(\frac{v_{i}\sin(\theta)}{g}\right)^2+y_0}$

$\color{#000000 }{ \displaystyle  {\rm Range}=\frac{v_{i}^{~2}\sin^2(\theta)}{g} -\frac{1}{2}\frac{v_{i}^{~2}\sin^2(\theta)}{g}+y_0}$

$\color{#000000 }{ \displaystyle  {\rm Range}=\frac{1}{2}\frac{v_{i}^{~2}\sin^2(\theta)}{g}+y_0}$

$\color{#000000 }{ \displaystyle  {\rm Range}=\frac{v_{i}^{~2}\sin^2(\theta)}{2g}+y_0}$

(this is the vertical range derivation)

solomonzelman · Answer

Horizontal Range derivation

$\color{#000000 }{ \displaystyle v_x(t)=v_{i}\cos(\theta)  }$

$\color{#000000 }{ \displaystyle \int v_x(t)~dt=\int v_{i}\cos(\theta)~dt  }$

$\color{#000000 }{ \displaystyle x(t)=v_{i}\cos(\theta)t}$

(wil omit the constant, or the initial position in the x, since we always choose $x_0=0$, by projectile motion. at least, I don't recal when this wasn't the case)

time to reach the top is (previously derived)
$\color{#000000 }{ \displaystyle t_{\rm top}=\frac{v_{i}\sin(\theta)}{g}}$

projectile path is symmetric about vertex/top, so

$\color{#000000 }{ \displaystyle t_{\rm ~flight}=t_{\rm ~top}=\frac{2v_{i}\sin(\theta)}{g}}$

Horizontal range,

$\color{#000000 }{ \displaystyle {\rm R_x}=x(\frac{2v_{i}\sin(\theta)t}{g})=v_{i}\cos(\theta)\frac{2v_{i}\sin(\theta)t}{g}}$

$\color{#000000 }{ \displaystyle {\rm R_x}=\frac{2v_{i}^{~2}\sin(\theta)\cos(\theta)}{g}=\frac{v_{i}^{~2}\sin(2\theta)}{g}}$

solomonzelman · Answer

and given horizontal or vertical range with some other information we can solve for initial velocity via this information