Question

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

Answer 1

please post the formula which is to be changed

Answer 2

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

Answer 3

y = ut sinQ - 1/2 g t
x = ut cosQ

i thought this was the formula, how did you get that?

Answer 4

no thats uniform motion
resnick book equation 4-25

Answer 5

trajectory of the path...

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

v0=u another name for initial velocity

Answer 12

just curious, but does

change trajectory of the path formula to make v0 the subject

just mean "solve for v0"?

Answer 13

yes

Answer 14

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

Answer 15

so it is algebra you need right?

Answer 16

yes .Sam but v is initial velocity

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

step by step would be good

Answer 22

:)

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

horizontal has only one equation v=d/t

Answer 27

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

Answer 28

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

Answer 29

do you have the projectile question?

Answer 30

no im just trying to understand the theory

Answer 31

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

Answer 32

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

Answer 33

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....

Answer 34

i did that but it doesn't quite change over

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

thanks