Solving y= (tan θ)x - (g/(2vi2cos2θ))x2

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In summary, to solve for θ, the given equation must be rewritten in terms of tanθ. This can be done by either expressing tanθ in terms of cosθ or cosθ in terms of tanθ. Once the equation has been rewritten, it can be solved using the quadratic formula.
  • #1
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Homework Statement


make [tex]\theta[/tex]the subject
[tex] y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2 [/tex]


Homework Equations





The Attempt at a Solution


not sure how to get start :(
should i change tan to sin/cos first?
 
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  • #2
Welcome to PF!

suppy123 said:

Homework Statement


make [tex]\theta[/tex]the subject
[tex] y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2 [/tex]

not sure how to get start :(
should i change tan to sin/cos first?

Hi suppy123! Welcome to PF! :smile:

You ultimately want to find an equation beginning "θ =".

But the last-step-but-one will be an equation beginning "cosθ =" or "tanθ =".

So the technique you need is to change the given equation so that it either has only cosθ (and no tanθ) or only tanθ (and no cosθ).

In other words: either rewrite tanθ in terms of cosθ, or cosθ in terms of tanθ (whichever seems easier). :smile:
 
  • #3
hi,
ok,
[tex]y = x(\frac{\sin{\theta}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{1}{\cos^2{\theta}})[/tex]

[tex]y = x(\frac{\cos{\theta+\frac{\pi}{2}}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{2}{1+\cos{2{\theta}}})[/tex]

is this correct? :S
 
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  • #4
Nooo …

You need the whole of "cosθ", not just the "cos" part.

An equation with both cosθ and cos(π/2 - θ) won't do. :frown:

Hint: tanθ is easier. How can you express 1/cos²θ in terms of tanθ? :smile:
 
  • #5
yup, it's [tex]\tan^2{\theta}+1[/tex]
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]
[tex]\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}[/tex]
[tex]\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}[/tex]

so
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{2\tan{\theta}}{\tan{2\theta}}+2)[/tex]
 
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  • #6
suppy123 said:
yup, it's [tex]\tan^2{\theta}+1[/tex]

ok, so the original equation in terms of tanθ is … ? :smile:
 
  • #7
is that right?
 
  • #8
sry it should be
[tex]\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1[/tex]

[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)[/tex]
 
  • #9
suppy123 said:
sry it should be
[tex]\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1[/tex]

[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)[/tex]

oh nooo … you've gone completely off the rails … :cry:

It's [tex]\tan^2{\theta} = \frac{-2\tan{\theta}}{\sin{2\theta}}+1[/tex]

Your original equation 1/cos²θ = tan²θ +1 was right.

But don't start using tan2θ … having both tan2θ and tanθ is as bad as having both cosθ and cos(π/2 - θ). :frown:

You want an equation with only tanθ, tan²θ, tan³θ, and so on. :smile:
 
  • #10
how about this?
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}+1)(\tan{\theta}-1)+2[/tex]
 
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  • #11
suppy123 said:
how about this?
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}-\sqrt{1})^2+1[/tex]

ooo … :cry:

I've just noticed that you did produce the right equation (before you went berserk):
suppy123 said:
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]

That's the equation you want!

Now that's a quadratic equation in tanθ, so use the usual -b ± √etc formula, to get an equation starting "tanθ =" :smile:
 
  • #12
lol...sweet, so
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]
[tex]0 = (-\frac{gx^2}{2vi^2})(\tan^2{\theta}) + x(\tan{\theta}) - (\frac{gx^2}{2vi^2} - y)[/tex]
[tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2}-y)^2}}{-2\frac{gx^2}{2vi^2-y}}[/tex]
 
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  • #13
oops, forgot to change the y heh
 
  • #14
suppy123 said:
[tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2})^2}}{-2\frac{gx^2}{2vi^2}}[/tex]

erm …
:cry: What happened to poor little y ? :cry:
 
  • #15
there you go
[tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)^2}}{-2(\frac{gx^2}{2vi^2})}[/tex]
;0
 
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  • #16
:biggrin: ooh … there he is … ! :biggrin:

hmm … that last equation doesn't look right, though.
 
  • #17
should be right this time ;)
[tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)}}{2(-\frac{gx^2}{2vi^2})}[/tex]
 
  • #18
:biggrin: Woohoo! :biggrin:

And now tidy it up a bit (get rid of some of the fractions)! :smile:
 

1. What is the formula for "Solving y= (tan θ)x - (g/(2vi2cos2θ))x2"?

The formula for solving this equation is y= (tan θ)x - (g/(2vi2cos2θ))x2, where θ represents the angle of elevation, g is the acceleration due to gravity, vi is the initial velocity, and x is the horizontal distance.

2. How do you solve for "x" in this equation?

To solve for x, you can use the quadratic formula or factor out the x term and solve for x using the zero product property.

3. What is the significance of the term (tan θ)x in this equation?

The term (tan θ)x represents the vertical displacement of the projectile due to the angle of elevation θ. It shows how high the projectile will go in the y direction.

4. How does the initial velocity affect the solution to this equation?

The initial velocity, represented by vi, affects the horizontal distance traveled by the projectile. The higher the initial velocity, the farther the projectile will travel in the x direction.

5. Can this equation be used to solve for the angle of elevation?

Yes, this equation can be rearranged and solved for the angle of elevation θ. However, it is recommended to use other equations specifically designed for finding the angle of elevation, such as the inverse tangent function.

Suggested for: Solving y= (tan θ)x - (g/(2vi2cos2θ))x2

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