MHB Kinematics + Trigonometry Application (Physics 12u)

[Wild ownz al]

Show that the maximum range of a projectile launched with Vi from Yi=0 is 45 degrees. (HINT: sin20 = 2sinθcosθ)

 

[MarkFL]

Show that the maximum range of a projectile launched with Vi from Yi=0 is 45 degrees. (HINT: sin20 = 2sinθcosθ)

Can you show what you have so far?

 

[Wild ownz al]

Can you show what you have so far?

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.

 

[MarkFL]

I would begin with velocity, with our coordinate axes oriented such that the initial position of the projectile is at the origin, and observe we are neglecting drag:

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

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

Hence:

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

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

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 \(t\)?

 

[Wild ownz al]

I would begin with velocity, with our coordinate axes oriented such that the initial position of the projectile is at the origin, and observe we are neglecting drag:

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

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

Hence:

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

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

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 \(t\)?

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

 

[MarkFL]

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

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:

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

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?

 

[Wild ownz al]

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:

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

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?

But I'm not looking for the distance traveled, I need to prove why 45 degrees is the max range.

 

[MarkFL]

But I'm not looking for the distance traveled, I need to prove why 45 degrees is the max range.

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.

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.

 

[MarkFL]

Let's find the root we want:

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

And so the range of the projectile is:

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

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

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

Can you proceed?

 

[Wild ownz al]

Let's find the root we want:

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

And so the range of the projectile is:

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

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

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

Can you proceed?

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

 

[MarkFL]

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

We will assume that $$\frac{v_0^2}{g}$$ 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:

$$\sin(2\theta)$$

In a problem like this, we will typically take:

$$0^{\circ}\le\theta\le90^{\circ}$$

This implies:

$$0^{\circ}\le2\theta\le180^{\circ}$$

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

 

[Wild ownz al]

We will assume that $$\frac{v_0^2}{g}$$ 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:

$$\sin(2\theta)$$

In a problem like this, we will typically take:

$$0^{\circ}\le\theta\le90^{\circ}$$

This implies:

$$0^{\circ}\le2\theta\le180^{\circ}$$

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

90 degrees? So theta = 45 degrees

 

[MarkFL]

90 degrees? So theta = 45 degrees

Yes, that's correct. :)

 

[Wild ownz al]

I would begin with velocity, with our coordinate axes oriented such that the initial position of the projectile is at the origin, and observe we are neglecting drag:

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

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

Hence:

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

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

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 \(t\)?

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.

 

[Wild ownz al]

We will assume that $$\frac{v_0^2}{g}$$ 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:

$$\sin(2\theta)$$

In a problem like this, we will typically take:

$$0^{\circ}\le\theta\le90^{\circ}$$

This implies:

$$0^{\circ}\le2\theta\le180^{\circ}$$

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

Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?

 

[MarkFL]

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.

We could set:

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

Let's multiply by -2:

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

Now, for the quadratic formula, we identify:

$$a=g$$

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

$$c=0$$

Hence:

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

And so we see our roots are:

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

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

Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?

We are assuming $$\frac{v_0^2}{g}$$ 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 \(\theta\).

 

[Wild ownz al]

we could set:

[math]y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t=0[/math]

let's multiply by -2:

[math]gt^2-2v_0\sin(\theta)t=0[/math]

now, for the quadratic formula, we identify:

[math]a=g[/math]

[math]b=-2v_0\sin(\theta)[/math]

[math]c=0[/math]

hence:

[math]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}[/math]

and so we see our roots are:

[math]y\in\left\{0,\frac{2v_0\sin(\theta)}{g}\right\}[/math]

factoring is much easier, and i recommend using it whenever you can.
We are assuming [math]\frac{v_0^2}{g}[/math] 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 \(\theta\).

thank you so much