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**rkman****Member**- Registered: 2005-10-02
- Posts: 5

If an object is projected at a fixed speed of u at a variable angle theta, how would one find out the equation of the curve that shows the furthest point the trajectory can be?

So if the theta= 90, then the body goes to it's highest possible point and is also where the curve would cut the y-axis. When theta = 45, the x value is v2/g (the maximum range) and shows where the curve cuts the x axis. So what is the equation of the curve that shows the maximum position the trajectory can travel to for any value of theta.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,659

For reference:

x = vt cos θ

y = vt sin θ - ½ g t²

You can find when it returns to it's original height by setting y=0:

y = vt sin θ - ½ g t² = 0

rearranging: vt sin θ = ½ g t²

divide by t: v sin θ = ½ g t

solving for t: t = v sin θ / ½ g = 2(v sin θ)/g

that is the **time** when it hits the ground, so it tells you **when** the max "x" value is:

max x = v cos θ × 2(v sin θ)/g = (2v²/g) cos θ sin θ

More neatly:

Now, are you asking for the maximum

a) x (horizontal distance), which I have just shown, or

b) y (height), or

c) actual distance from origin, which would be √(x²+y²)

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**rkman****Member**- Registered: 2005-10-02
- Posts: 5

I know how to find the maximum range and the maximum height. I also know the trajectory equation which will tell me who to find the displacement from the origin.

What I want to know, and it's hard to put, is the equation for the maximum point of displacements for a fixed velocity v, but a variant angle.

Imagine you draw out the axis, and set u= 100. Then for theta=1, you draw the path. Then for theta=2,...every single value of theta from 0 to 90. If done by a computer, the final image produced would look like a -x^2 graph with the area shaded. I want to know the equation of that curve given i.e. the maximum displacement from the origin.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,659

OK, so you want a formula like:

d_max = f(θ)

But I still don't know if displacement is in terms of x, y or distance from origin (the hardest!), so I will just demonstrate with x. As I showed before:

x_max = (2v²/g) cos θ sin θ

Now, we are lucky because the double-angle formula for sin is: sin 2θ = 2 sin θ cos θ, so we can substitute:

x_max = (v²/g) sin 2θ

And because you have fixed "v" and "g" (gravity), they could be combined into, say: C = v²/g

x_max = C sin 2θ

And that is maximum range in terms of angle.

But you are now going to say that you want to know maximum actual distance, ie the maximum might be somewhere in the sky during it's path, is that right?

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**rkman****Member**- Registered: 2005-10-02
- Posts: 5

Above is an image of what i mean. The green shows random trajectory values for theta greater than 45. The blue line is the path for theta equals 45. The orange line is the path chosen by theta equal to 90 degrees (i.e. straight up and down the y-axis)

I want to find the equation of the red path. If i take any point on this red curve or the area below it, I could find a theta for the trajectory to reach this point. If I use a point exactly on this line then I'm only going to get one repeated solution for theta. And if I chose any point above the red line, then there are no solutions for theta since the magnitude of v is not great enough.

Hope that makes it clearer.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If I'm right, you want to find the value of θ that gives the highest value of y, in terms of v and x.

Start with the equation that you gave in the other trajectory topic:

y = x tanθ - gx²(1+ tan² θ)/2v²

We need to find the maximum of y by changing θ. This can be found by differentiating with respect to θ and solving for when that expression is equal to 0. dy/dx = x sec² θ - 2gx² (tan θsec² θ)/2v². [Note: sec θ = 1/cos θ]

The value of θ for which this expression is equal to 0 is the value which will give the highest value of y.

So, x sec² θ - 2gx² (tan θsec² θ)/2v²

Rearrange: x sec² θ = 2gx² (tan θsec² θ)/2v²

Cancel various things: 1 = (gx tan θ)/v²

Solve: θ = tan-¹ (v²/gx)

Happily, the only time θ appears in your original equation is as a function of tan, which means that the tan and tan-¹ will always cancel out.

Your final equation, after a bit of rearrangement, is y = 3v²/2g - gx²/2v², where v is the initial velocity, x is the distance from start point and g is acceleration due to gravity. The equation only works when it returns a value of y that is positive.

Why did the vector cross the road?

It wanted to be normal.

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