Kinematics + Trigonometry Application (Physics 12u)

In summary: I think I'm starting to see it now, but how do I proceed without knowing theta or...Theta is not required for this problem. What you need to do is find a value of \(v_0\) that results in the maximum range. Once you have found this, then you can determine the value of \(\theta\) that results in that range.
  • #1
Wild ownz al
30
0
Show that the maximum range of a projectile launched with Vi from Yi=0 is 45 degrees. (HINT: sin20 = 2sinθcosθ)
 
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  • #2
Wild ownz al said:
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?
 
  • #3
MarkFL said:
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.
 
  • #4
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:

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

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

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 \(t\)?
 
  • #5
MarkFL said:
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:

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

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

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 \(t\)?

Do I use the quadratic formula to determine that? Or is it 1/2 t?
 
  • #6
Wild ownz al said:
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:

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

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?
 
  • #7
MarkFL said:
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?

But I'm not looking for the distance traveled, I need to prove why 45 degrees is the max range.
 
  • #8
Wild ownz al said:
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.
 
  • #9
Let's find the root we want:

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

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?
 
  • #10
MarkFL said:
Let's find the root we want:

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

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?

I think I'm starting to see it now, but how do I proceed without knowing theta or Vi?
 
  • #11
Wild ownz al said:
I think I'm starting to see it now, but how do I proceed without knowing theta or Vi?

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?
 
  • #12
MarkFL said:
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?

90 degrees? So theta = 45 degrees
 
  • #13
Wild ownz al said:
90 degrees? So theta = 45 degrees

Yes, that's correct. :)
 
  • #14
MarkFL said:
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:

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

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

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 \(t\)?

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.
 
  • #15
MarkFL said:
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?

Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?
 
  • #16
Wild ownz al said:
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:

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

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.

Wild ownz al said:
Also how do you go from Vi^2 / g to 0 degrees < theta < 90 degrees?

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 \(\theta\).
 
  • #17
markfl said:
we 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 \(\theta\).

thank you so much
 

FAQ: Kinematics + Trigonometry Application (Physics 12u)

What is kinematics in physics?

Kinematics is the branch of physics that deals with the study of motion, including the position, velocity, and acceleration of objects without considering the forces that cause the motion.

How is trigonometry used in kinematics?

Trigonometry is used in kinematics to analyze the motion of objects in two or three dimensions. It helps to calculate the displacement, velocity, and acceleration of an object by using trigonometric functions such as sine, cosine, and tangent.

What is the difference between distance and displacement?

Distance is the total length of the path traveled by an object, while displacement is the shortest distance between the initial and final positions of the object. Displacement takes into account the direction of motion, while distance does not.

How is kinematics applied in real life?

Kinematics is applied in real life in various fields such as engineering, sports, and transportation. It is used to design structures, analyze the motion of athletes, and calculate the trajectory of objects in motion, such as airplanes and cars.

What are the three equations of motion in kinematics?

The three equations of motion in kinematics are:

  • v = u + at (velocity = initial velocity + acceleration x time)
  • s = ut + 1/2at² (displacement = initial velocity x time + 1/2 x acceleration x time squared)
  • v² = u² + 2as (velocity squared = initial velocity squared + 2 x acceleration x displacement)

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