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#### Wild ownz al

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- Nov 11, 2018

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- Thread starter Wild ownz al
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- Nov 11, 2018

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Can you show what you have so far?

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- #3

- Nov 11, 2018

- 30

So far...Can you show what you have so far?

Xf = Xi + Vixcosθt

Xf = 0 + Vixcosθt

Xf = Vixcosθt

t = Xf / Vixcosθ

Then I use position equation

Yf = 0 + ViySinθt+1/2ayt^2

Plug in t = Xf / Vixcosθ

Yf = 0 + ViySinθ(Xf/Vixcosθ)+1/2ay(Xf/Vixcosθ)^2

This is what I have so far, however I still don't understand HOW I am suppose to prove the max range with trig, though I do know we are suppose to use the hint given, thanks.

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\(\displaystyle v_x(t)=v_0\cos(\theta)\)

\(\displaystyle v_y(t)=-gt+v_0\sin(\theta)\)

Hence:

\(\displaystyle x(t)=v_0\cos(\theta)t\)

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=\frac{t}{2}\left(2v_0\sin(\theta)-gt\right)\)

Now, the range of the projectile is where it returns to the ground, which is the non-zero root of \(y(t)\), which is what value of \(t\)?

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- #5

- Nov 11, 2018

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Do I use the quadratic formula to determine that? Or is it 1/2 t?

\(\displaystyle v_x(t)=v_0\cos(\theta)\)

\(\displaystyle v_y(t)=-gt+v_0\sin(\theta)\)

Hence:

\(\displaystyle x(t)=v_0\cos(\theta)t\)

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=\frac{t}{2}\left(2v_0\sin(\theta)-gt\right)\)

Now, the range of the projectile is where it returns to the ground, which is the non-zero root of \(y(t)\), which is what value of \(t\)?

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- #6

I factored \(y(t)\) so you could more easily determine the roots. We see that one root is \(t=0\), and that corresponds to the moment when the projectile is launched from the ground. We don't want that root, we want the other one, corresponding to when it lands (when its y-coordinate is also zero), or returns to the ground. So, we equate the other factor to zero to determine when that is:Do I use the quadratic formula to determine that? Or is it 1/2 t?

\(\displaystyle 2v_0\sin(\theta)-gt=0\)

Solving this for \(t\) will tell us when the projectile lands, and then we can use this value of \(t\) in \(x(t)\) to determine how far (horizontally) is traveled, which is by definition, its range.

Can you proceed?

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- #7

- Nov 11, 2018

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But I'm not looking for the distance traveled, I need to prove why 45 degrees is the max range.I factored \(y(t)\) so you could more easily determine the roots. We see that one root is \(t=0\), and that corresponds to the moment when the projectile is launched from the ground. We don't want that root, we want the other one, corresponding to when it lands (when its y-coordinate is also zero), or returns to the ground. So, we equate the other factor to zero to determine when that is:

\(\displaystyle 2v_0\sin(\theta)-gt=0\)

Solving this for \(t\) will tell us when the projectile lands, and then we can use this value of \(t\) in \(x(t)\) to determine how far (horizontally) is traveled, which is by definition, its range.

Can you proceed?

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- #8

What you will obtain is the range in terms of the parameters \(v_0\) and \(\theta\). And then you will need to demonstrate that for some positive value of \(v_0\), which value of \(\theta\) will maximize the range.But I'm not looking for the distance traveled, I need to prove why 45 degrees is the max range.

In other words, in order to maximize the range, we need to know what that range is, in terms of our parameters. The range is our objective function.

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- #9

\(\displaystyle 2v_0\sin(\theta)-gt=0\implies t=\frac{2v_0\sin(\theta)}{g}\)

And so the range of the projectile is:

\(\displaystyle x_{\max}=x\left(\frac{2v_0\sin(\theta)}{g}\right)=v_0\cos(\theta)\left(\frac{2v_0\sin(\theta)}{g}\right)=\frac{2v_0^2\sin(\theta)\cos(\theta)}{g}\)

Applying the double-angle identity for sine, we obtain:

\(\displaystyle x_{\max}=\frac{v_0^2\sin(2\theta)}{g}\)

Can you proceed?

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- #10

- Nov 11, 2018

- 30

I think I'm starting to see it now, but how do I proceed without knowing theta or Vi?

\(\displaystyle 2v_0\sin(\theta)-gt=0\implies t=\frac{2v_0\sin(\theta)}{g}\)

And so the range of the projectile is:

\(\displaystyle x_{\max}=x\left(\frac{2v_0\sin(\theta)}{g}\right)=v_0\cos(\theta)\left(\frac{2v_0\sin(\theta)}{g}\right)=\frac{2v_0^2\sin(\theta)\cos(\theta)}{g}\)

Applying the double-angle identity for sine, we obtain:

\(\displaystyle x_{\max}=\frac{v_0^2\sin(2\theta)}{g}\)

Can you proceed?

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- #11

We will assume that \(\displaystyle \frac{v_0^2}{g}\) is some constant, and only worry about the factor that is a function of \(\theta\). Since it is in the numerator, we want to maximize:I think I'm starting to see it now, but how do I proceed without knowing theta or Vi?

\(\displaystyle \sin(2\theta)\)

In a problem like this, we will typically take:

\(\displaystyle 0^{\circ}\le\theta\le90^{\circ}\)

This implies:

\(\displaystyle 0^{\circ}\le2\theta\le180^{\circ}\)

On this interval, what value of \(2\theta\) maximizes the sine function, and hence, what value of \(\theta\) maximizes the sine function?

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- #12

- Nov 11, 2018

- 30

90 degrees? So theta = 45 degreesWe will assume that \(\displaystyle \frac{v_0^2}{g}\) is some constant, and only worry about the factor that is a function of \(\theta\). Since it is in the numerator, we want to maximize:

\(\displaystyle \sin(2\theta)\)

In a problem like this, we will typically take:

\(\displaystyle 0^{\circ}\le\theta\le90^{\circ}\)

This implies:

\(\displaystyle 0^{\circ}\le2\theta\le180^{\circ}\)

On this interval, what value of \(2\theta\) maximizes the sine function, and hence, what value of \(\theta\) maximizes the sine function?

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- #13

Yes, that's correct.90 degrees? So theta = 45 degrees

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- #14

- Nov 11, 2018

- 30

How could I have used the quadratic formula to determine the roots? My factoring is that great and it seems odd to do without given coefficients.

\(\displaystyle v_x(t)=v_0\cos(\theta)\)

\(\displaystyle v_y(t)=-gt+v_0\sin(\theta)\)

Hence:

\(\displaystyle x(t)=v_0\cos(\theta)t\)

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=\frac{t}{2}\left(2v_0\sin(\theta)-gt\right)\)

Now, the range of the projectile is where it returns to the ground, which is the non-zero root of \(y(t)\), which is what value of \(t\)?

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- #15

- Nov 11, 2018

- 30

Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?We will assume that \(\displaystyle \frac{v_0^2}{g}\) is some constant, and only worry about the factor that is a function of \(\theta\). Since it is in the numerator, we want to maximize:

\(\displaystyle \sin(2\theta)\)

In a problem like this, we will typically take:

\(\displaystyle 0^{\circ}\le\theta\le90^{\circ}\)

This implies:

\(\displaystyle 0^{\circ}\le2\theta\le180^{\circ}\)

On this interval, what value of \(2\theta\) maximizes the sine function, and hence, what value of \(\theta\) maximizes the sine function?

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- #16

We could set:How could I have used the quadratic formula to determine the roots? My factoring is that great and it seems odd to do without given coefficients.

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=0\)

Let's multiply by -2:

\(\displaystyle gt^2-2v_0\sin(\theta)t=0\)

Now, for the quadratic formula, we identify:

\(\displaystyle a=g\)

\(\displaystyle b=-2v_0\sin(\theta)\)

\(\displaystyle c=0\)

Hence:

\(\displaystyle y=\frac{-(-2v_0\sin(\theta))\pm\sqrt{(-2v_0\sin(\theta))^2-4(g)(0)}}{2g}=\frac{2v_0\sin(\theta)\pm2v_0\sin(\theta)}{2g}=\frac{v_0\sin(\theta)\pm v_0\sin(\theta)}{g}\)

And so we see our roots are:

\(\displaystyle y\in\left\{0,\frac{2v_0\sin(\theta)}{g}\right\}\)

Factoring is much easier, and I recommend using it whenever you can.

We are assuming \(\displaystyle \frac{v_0^2}{g}\) is some arbitrary positive constant. Since it remains fixed, we need only concern ourselves with the factor that will vary in accordance with the parameter in which we're interested, which is \(\theta\).Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?

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- #17

- Nov 11, 2018

- 30

thank you so muchwe could set:

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=0\)

let's multiply by -2:

\(\displaystyle gt^2-2v_0\sin(\theta)t=0\)

now, for the quadratic formula, we identify:

\(\displaystyle a=g\)

\(\displaystyle b=-2v_0\sin(\theta)\)

\(\displaystyle c=0\)

hence:

\(\displaystyle y=\frac{-(-2v_0\sin(\theta))\pm\sqrt{(-2v_0\sin(\theta))^2-4(g)(0)}}{2g}=\frac{2v_0\sin(\theta)\pm2v_0\sin(\theta)}{2g}=\frac{v_0\sin(\theta)\pm v_0\sin(\theta)}{g}\)

and so we see our roots are:

\(\displaystyle y\in\left\{0,\frac{2v_0\sin(\theta)}{g}\right\}\)

factoring is much easier, and i recommend using it whenever you can.

We are assuming \(\displaystyle \frac{v_0^2}{g}\) is some arbitrary positive constant. Since it remains fixed, we need only concern ourselves with the factor that will vary in accordance with the parameter in which we're interested, which is \(\theta\).