change trajectory of the path formula to make v0 the subject...
please do this step by step

- anonymous

change trajectory of the path formula to make v0 the subject...
please do this step by step

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- experimentX

please post the formula which is to be changed

- anonymous

y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2))

- experimentX

y = ut sinQ - 1/2 g t
x = ut cosQ
i thought this was the formula, how did you get that?

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## More answers

- anonymous

no thats uniform motion
resnick book equation 4-25

- anonymous

trajectory of the path...

- anonymous

apparently its the horizontal motion and vertical motion put together....but i don't understand how they did that

- experimentX

I understand what you are doing ... in my first equation, we have variable t,
you get your equation by replacing t with x from second variable

- experimentX

*** y = ut sinQ - 1/2 g t ^2

- experimentX

and what do you mean by vo as subject .. is v. instantaneous velocity??

- anonymous

yeh i kinda figured that a bit but then it still doesn't quite fit

- anonymous

v0=u another name for initial velocity

- anonymous

just curious, but does
change trajectory of the path formula to make v0 the subject
just mean "solve for v0"?

- anonymous

yes

- .Sam.

\[y=\tan \theta x-(\frac{gx^2}{(2vcos(x))^2})\] Is it?

- anonymous

so it is algebra you need right?

- anonymous

yes .Sam but v is initial velocity

- .Sam.

\[v=\sqrt{\frac{gx^2}{4\cos^{2}x(\tan \theta x-y)}}\]

- anonymous

yes and also how they put the equations together...the vertical motion and horizontal motion...

- anonymous

\[y=x\tan (\theta) -\frac{gx^2}{2v_0cos^2(x)^2}\]

- anonymous

or maybe i have it wrong and it is what .sam. wrote

- anonymous

step by step would be good

- anonymous

:)

- experimentX

x and y is a particular point on your trajectory, theta is the angle of projection

- experimentX

g is accn due to gravity, you have all parameters you need to calculate u

- .Sam.

There are some kinematic equations you can use,
\[v_{f}=v_{i}+at\]
\[v_{f}^2=v_{i}^2+2as\]

- .Sam.

horizontal has only one equation v=d/t

- anonymous

how does those two equations make the trajectory of the path equation?

- anonymous

y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2))
\[y=(\tan \theta )x -gx^2/2(v0\cos \theta)^2 \]

- .Sam.

do you have the projectile question?

- anonymous

no im just trying to understand the theory

- experimentX

x, y givies you the path of the projectile ... thats parabolic

- .Sam.

because things mixed up pretty bad here y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2))

- anonymous

it says equation x-x0=(v0cos theta)t and y-y0=v0sin thetat-1/2gt^2 make the trajectory equation by replacing t with x....

- anonymous

i did that but it doesn't quite change over

- .Sam.

So the idea is removing t from your new equation,
x-x0=(v0cos theta)t .................(1)
y-y0=v0sin thetat-1/2gt^2.......(2)
From (1),
\[ \huge t=\frac{x-x_{i}}{v_{i}\cos \theta}\]
Sub (t) into (2),
\[y-y_{i}=v_{i}\sin \theta t-\frac{1}{2}g t^2\]
\[y-y_{i}=v_{i}\sin \theta (\frac{x-x_{i}}{v_{i}\cos \theta})-\frac{1}{2}g(\frac{x-x_{i}}{v_{i}\cos \theta})^2\]

- .Sam.

Did they say something like "solve in terms of" ?

- anonymous

ahh i see it now....got mixed up with the change in x and y...
well keep all the values there...i just want v0 to be the subject

- anonymous

so i can sub all the values in
sorry for slow reply, internet is playing up.

- anonymous

because if its just y it is not that useful...the equation is so big

- .Sam.

\[\frac{1}{2}g(\frac{x-x_{i}}{v_{i}\cos \theta})^2=\tan \theta(x-x_{i})-(y-y_{i})\]
\[g(\frac{x-x_{i}}{v_{i}\cos \theta})^2=2[\tan \theta(x-x_{i})-(y-y_{i})]\]
\[\frac{(x-x_{i})^2}{v_{i}^2\cos^2 \theta}=\frac{2[\tan \theta(x-x_{i})-(y-y_{i})]}{g}\]
\[\frac{1}{v_{i}^2}=\frac{2[\tan \theta(x-x_{i})-(y-y_{i})]\cos^2 \theta}{g(x-x_{i})^2}\]
\[v_{i}^2=\frac{g(x-x_{i})^2}{2[\tan \theta(x-x_{i})-(y-y_{i})]\cos^2 \theta}\]
\[v_{i}=\sqrt{\frac{g(x-x_{i})^2}{2[\tan \theta(x-x_{i})-(y-y_{i})]\cos^2 \theta}}\]

- anonymous

ahhh i see what i did wrong....accidently forgot the square of cos^2 theta

- anonymous

thanks

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