Would someone be able to isolate v?

[terff]

Homework Statement

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)

Homework Equations

N/A

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.

 

[Simon Bridge]

1. Homework Statement
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)

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

Well my attempt was,

∆y = v₀(sinθ)(∆x/v₀(cosθ)) + ½(a)(∆x/v₀(cosθ))²

... 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;

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.

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

 

[Ray Vickson]

Homework Statement

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)

Homework Equations

N/A

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.

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)}}

 

[terff]

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?

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
instead I was getting 8.56

 

[Simon Bridge]

Ahh my apologies,

$$\Delta y = v(\sin\theta)\left(\frac{\Delta x}{v_0\cos\theta}\right) + \frac{1}{2}(a)\left(\frac{\Delta x}{v_0(\cos\theta)^2}\right)$$
is the equation, and yes that is what I wrote.

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

 

[terff]

... check that equation by dimensional analysis.

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

 

[Simon Bridge]

I got 6.4104

...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.