anonymous
  • anonymous
change trajectory of the path formula to make v0 the subject... please do this step by step
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
please post the formula which is to be changed
anonymous
  • anonymous
y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2))
experimentX
  • experimentX
y = ut sinQ - 1/2 g t x = ut cosQ i thought this was the formula, how did you get that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
no thats uniform motion resnick book equation 4-25
anonymous
  • anonymous
trajectory of the path...
anonymous
  • anonymous
apparently its the horizontal motion and vertical motion put together....but i don't understand how they did that
experimentX
  • 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
  • experimentX
*** y = ut sinQ - 1/2 g t ^2
experimentX
  • experimentX
and what do you mean by vo as subject .. is v. instantaneous velocity??
anonymous
  • anonymous
yeh i kinda figured that a bit but then it still doesn't quite fit
anonymous
  • anonymous
v0=u another name for initial velocity
anonymous
  • anonymous
just curious, but does change trajectory of the path formula to make v0 the subject just mean "solve for v0"?
anonymous
  • anonymous
yes
.Sam.
  • .Sam.
\[y=\tan \theta x-(\frac{gx^2}{(2vcos(x))^2})\] Is it?
anonymous
  • anonymous
so it is algebra you need right?
anonymous
  • anonymous
yes .Sam but v is initial velocity
.Sam.
  • .Sam.
\[v=\sqrt{\frac{gx^2}{4\cos^{2}x(\tan \theta x-y)}}\]
anonymous
  • anonymous
yes and also how they put the equations together...the vertical motion and horizontal motion...
anonymous
  • anonymous
\[y=x\tan (\theta) -\frac{gx^2}{2v_0cos^2(x)^2}\]
anonymous
  • anonymous
or maybe i have it wrong and it is what .sam. wrote
anonymous
  • anonymous
step by step would be good
anonymous
  • anonymous
:)
experimentX
  • experimentX
x and y is a particular point on your trajectory, theta is the angle of projection
experimentX
  • experimentX
g is accn due to gravity, you have all parameters you need to calculate u
.Sam.
  • .Sam.
There are some kinematic equations you can use, \[v_{f}=v_{i}+at\] \[v_{f}^2=v_{i}^2+2as\]
.Sam.
  • .Sam.
horizontal has only one equation v=d/t
anonymous
  • anonymous
how does those two equations make the trajectory of the path equation?
anonymous
  • anonymous
y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2)) \[y=(\tan \theta )x -gx^2/2(v0\cos \theta)^2 \]
.Sam.
  • .Sam.
do you have the projectile question?
anonymous
  • anonymous
no im just trying to understand the theory
experimentX
  • experimentX
x, y givies you the path of the projectile ... thats parabolic
.Sam.
  • .Sam.
because things mixed up pretty bad here y=tan(theta)x -( (gx^2)/ (2(v0cos(theta))^2))
anonymous
  • 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
  • anonymous
i did that but it doesn't quite change over
.Sam.
  • .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.
  • .Sam.
Did they say something like "solve in terms of" ?
anonymous
  • 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
  • anonymous
so i can sub all the values in sorry for slow reply, internet is playing up.
anonymous
  • anonymous
because if its just y it is not that useful...the equation is so big
.Sam.
  • .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
  • anonymous
ahhh i see what i did wrong....accidently forgot the square of cos^2 theta
anonymous
  • anonymous
thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.