# Increasing distance

I'm having trouble getting the answer without doing a guess and check.

Q: Find the maximum angle a projectile can be shot so that it's distance from the origin is always increasing. Air resistance negleced.

So there is the x and y position based on initial velocity, angle, and time, but I made initial velocity equal 1 for simplicity. Next, make a vector out of this and find the distance of the vector at varying angle and time.

Take the time derivative and make sure it is never negative. Unfortunately, there is no angle where change in the projectile's distance is zero at the point where it lands, so I can not make the whole equation all in terms of the angle.

So what can I do?

I'm pretty damn sure the answer is ~70.356 deg.

Tide
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I think you'll find that the distance the projectile travels is proportional to $$sin\theta \times cos\theta$$. For what angle is that a maximum?

HallsofIvy
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well, I just graphed it over and over using a calculator changing the angle until I got really really close to a non posative slope.

$$sin\theta \times cos\theta$$
has a maximum at 45 deg.

Position:
$$x = v_{0x}t = v cos(\theta)t$$
$$y = v_{0y}t -.5gt^2 = v sin(\theta)t - .5gt^2$$

Let $\vec {C}$ be the vector from the origin to the position of projectile

$$\vec {C} = [x,y] = [v cos(\theta)t,v sin(\theta)t - .5gt^2]$$

$$||\vec {C}|| = \sqrt{x^2 + y^2} = \sqrt{(v cos(\theta)t)^2+(v sin(\theta)t - .5gt^2)^2}$$

Take the time derivative and find where it equals zero. (let v=1 for simplicity)
$$\frac {d||\vec {C}||} {dt} =\frac {2t + 4(4.9)^2t^3 - 3(9.8)sin(\theta)t^2} {\sqrt{( cos(\theta)t)^2 + (sin(\theta)t - .5gt^2)^2}}$$

But I can't do that without guessing and checking. I hoped to find the derivative of C to be zero at the landing point, so that I could sub all the t's for
$$\frac {sin(\theta)} {4.9}$$

but no luck with that one.

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Tide
Homework Helper
HallsofIvy said:

If the projectile is launched from the origin (0, 0) then with x -> horizontal and y -> vertical we have $$x = v_{x0} t$$ and $$y = v_{y0} t - \frac{1}{2} g t^2$$. Here $$v_{x0} = v_0 cos \theta$$ and $$v_{y0} = v_0 sin \theta$$.

The projectile will return to the ground at time $$t = T = \frac{2v_{y0}}{g}$$ at which time its horizontal displacement (range) will be $$x(T) = 2* sin\theta cos\theta v_0^2 / g$$.

By the way, "damn" is a verb - I am sure you meant to use the adjective or adverb form "damned!"

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Well, you got a function of theta only, but I'm not so sure this helps me. I am not trying to find an angle that will give me maximum range, but a maximum angle that will yield increasing distance from the origin.

Tide
Homework Helper
Make a graph of $$\sin \theta \times \cos \theta$$ and tell me what you see! :-)

How do I do a cross product of two scalars? I'm sure that's not what you ment. So, how does this help me?

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Tide
Homework Helper
It's not a cross product - it's scalar multiplication. That's a regular multiplication symbol! I'm not fully accustomed to this LaTeX system yet!

Ok, so how does that help?
$$\sin \theta \cos \theta$$

is a function of maximum x-displacement.

The attached image shows the vector C that must never decrease in magnatude as the projectile moves along the arc.

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hmm perhaps something to do with the tangent must not be perpendicular or turned to the right of C at time t

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Tide
Homework Helper
Thanks for the picture. It made me realize I was locked onto the wrong problem - i.e. it has nothing to do with range!

Here's what I tried - in outline form since the algebra gets to be a bit heavy: You can explicitly write the time dependent displacement (x, y) and velocity as vectors. Now form the scalar product of those two quantities. The result is proportional to the cosine of the angle that the position vector makes with the velocity vector.

As long as it is positive, the projectile is moving away from the origin. It's a complicated function of time but your objective is simply to determine when it is positive. Factor out from it everything that you can and recognize that time (t) is always nonnegative. You will end up with a quadratic function of time It is helpful to complete the square and, recognizing that the graph of the function is a parabola, you can determine the axis of the parabola. Basically you want a parabola that has a minimum at its vertex and the vertex must lie above the t axis. This places a constraint on $$\theta$$ and I finally arrived at the expression $$\sin \theta < \sqrt{\frac{8}{9}}$$ so the critical angle is $$\arcsin {\sqrt \frac{8}{9}}$$ which is pretty close to what you estimated in your original post.

I am still not understanding. I completed the squares on $\vec {C} \dotprod {\frac {d \vec {C}} {dt}} = 0$ and get this really complex function that tells me nothing.
$$.5g^2[(t - \frac {3 V_0 sin \theta} {2g})^2 + (2 - \frac {9} {4} sin^2 \theta) \frac {V^2_0} {g^2}] = 0$$