# Would someone be able to isolate v?

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1. Nov 10, 2014

### terff

1. The problem statement, all variables and given/known data
So I have a group lab for physics, but I need some alg/trig help isolating v for a derived equation.
I've tried many times to to get v0 by itself but when I check the equation with the values that I have for each variable I always get the wrong answer.
∆y = v0(sinθ)(∆x/v0(cosθ)) + ½(a)(∆x/v0(cosθ))2)
2. Relevant equations
N/A
3. The attempt at a solution
Well my attempt was,
∆y = v0(sinθ)(∆x/v0(cosθ)) + ½(a)(∆x/v0(cosθ))2
then simplify the fractions in the addition
∆y = (sinθ)∆x/cosθ + (a)(∆x2)/(2v02(cosθ)2)
and then multiply by cosθ2 on both sides
∆y(cosθ)2 = (cosθ)(sinθ)(∆x) + (a)(∆x2)/2v02
subtract (cosθ)(sinθ)(∆x) on both sides
∆y(cosθ)2 - (cosθ)(sinθ)(∆x) = (a)(∆x2)/2v02
multiply 2v02 on both sides, then divide by 2∆y(cosθ)2 - (cosθ)(sinθ)(∆x) to get
v02 by itself
(a)(∆x2)/(2∆y(cosθ)2 - (cosθ)(sinθ)(∆x)) = v02
and then sqrt both sides
√(a(∆x2)/(2∆y(cosθ)2 - (cosθ)(sinθ)(∆x))) = v0

I tried it a few times with known values, once it came back correct, but with the other values, I tried multiple times and always got a wrong answer.

Last edited: Nov 10, 2014
2. Nov 10, 2014

### Simon Bridge

So this equation, and you want to put it in the form $v_0=\cdots$

... this first line is different from what you wrote above. Which is it?

Continuing form here, am I reading that correctly? You wrote:
$$\Delta y = v_0\sin\theta\left(\frac{\Delta x}{v_0\cos\theta}\right) + \frac{1}{2}a\left( \frac{\Delta x}{v_0\cos\theta} \right)^2$$
... that looks likely since it appears to be a ballistics calculation with $\Delta x/(v_0\cos\theta)$ being the time of flight, and a will end up being -g.

This simplifies to:
$$2 (\Delta y \cos^2\theta)v_0^2 = (2 \Delta x \sin\theta\cos\theta)v_0^2 + a(\Delta x)^2$$ ... which seems straight forward enough;

... how did you know you got the wrong answer?

3. Nov 10, 2014

### Ray Vickson

I will write v instead of v0, and x, y instead of Δx, Δy. So, your equation is
$$y = \frac{v \sin(\theta) x}{v \cos(\theta)} + \frac{a}{2} \left( \frac{x}{v \cos(\theta)} \right)^2 = x\tan(\theta)+ \frac{a x^2}{2 v^2 \cos(\theta)^2}$$
The solution is straightforward:
$$v = \pm \frac{1}{\sqrt{2}} \left| \frac{x}{\cos(\theta)}\right| \sqrt{ \frac{a}{y - x \tan(\theta)}}$$

Last edited: Nov 10, 2014
4. Nov 10, 2014

### terff

Ahh my apologies,
$∆y = v(sinθ)(\frac{∆x}{v0(cosθ)}) + \frac {1}{2}(a)(\frac{∆x}{v0(cosθ)^2})$
is the equation, and yes that is what I wrote.

In my final equation
$\sqrt{\frac{2a∆x}{2∆y(cosθ)^2 - (cosθ)(sinθ)(∆x)}} = v$
substituted the following variables
θ angle= 60º
∆x= 3.78 m
∆y= -.268 m
ay= -9.8 m/s2
$\sqrt{\frac{2(-9.8)(3.78)}{2(-.268)(cos60º)^2 - (cos60º)(sin60º)(3.78)}} = v$
and I was supposed to get 6.87 m/s for v

5. Nov 10, 2014

### Simon Bridge

... That equation is incorrect: check by dimensional analysis and redo the derivation.
(Assuming the leading "v" is a typo and should be $v_0$?)

Last edited: Nov 10, 2014
6. Nov 10, 2014

### terff

oops I realize another mistake I did in that equation $$\Delta y = v_0\sin\theta\left(\frac{\Delta x}{v_0\cos\theta}\right) + \frac{1}{2}a\left( \frac{\Delta x}{v_0\cos\theta} \right)^2$$ is correct, the squared is in the wrong place in my fraction.

$-0.268 = v(0.866)(\frac{3.78}{v(0.5)}) + \frac {1}{2}(-9.8)(\frac{3.78}{v(0.5)})^2$

I got 6.4104

7. Nov 10, 2014

### Simon Bridge

...m/s yep real physicists get fussy about units too.

Always state the units somewhere.

It is likely the equation is correct now - I cannot be sure because I don't know the experiment you did.
Where does the "correct" value come from? How do you know it's correct?

It looks like you have substituted the values in too soon though ... do the algebra first.

As a rule - the values you measure in the experiment are "correct".
They are the values you got - unless someone wants to say you are lying. Maybe you made a mistake in measurement, maybe you rounded off too far or failed to account for some variation in an instrument reading? If your values disagree with some theoretical value, then your conclusion will state that your results did not support the theory... but you do want to rule out some silly rounding or arithmetic error first.