# Optimal Angle

1. Aug 18, 2005

### bomba923

I shoot a cannon from a cliff of height $h$, with an initial velocity $v_0$ and angle of elevation $$\theta$$. If $h > 0$, what $$\theta$$ will maximize the cannon's range (how far the cannonball lands from base of the cliff), assuming no air resistance?
Ok...
The cannonball's height at any moment within its trajectory is
$$y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2}$$

The cannon will land at ground zero, so I find that the time spent in the air is:
$$y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2} = 0 \Rightarrow t = \frac{{v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{g}$$

And therefore the range is:
$$x = v_0 \frac{{v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{g} \cos \theta = \frac{{v_0 }}{g}\left( {v_0 \sin \theta \cos \theta + \left( {\cos \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} } \right)$$

*But, which $$\theta$$ will maximize the range? Here, I simply find the roots of ${dx}/{d\theta}$:
$$\frac{{dx}}{{d\theta }} = \frac{{v_0 }}{g}\left[ {v_0 \cos 2\theta + \frac{{v_0^2 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }} - \left( {\sin \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} } \right] = 0 \Rightarrow$$
$$v_0 \cos 2\theta \; + \; \frac{{v_0^2 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }} \; - \; \left( {\sin \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} = 0 \Rightarrow$$

To simplify this equation, I can multiply both sides by
$$\frac{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{{v_0 \sin \theta }}$$
and continue to get :

$$\left( {\cos 2\theta } \right)\sqrt {v_0^2 + 2gh\csc ^2 \theta } \; + \; v_0 \cos 2\theta \; + \; 2v_0^{ - 1} gh = 0 \Rightarrow - \sqrt {v_0^2 + 2gh\csc ^2 \theta } = v_0 \; + \; 2v_0^{ - 1} gh\sec 2\theta \Rightarrow$$

$$\csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right)$$

Now, we can multiply both sides by $\cos 2\theta$, to get
$$\cot ^2 \theta - 1 = 2\left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) \Rightarrow \cot ^2 \theta - 2v_0^{ - 2} gh\sec 2\theta = 3 \Rightarrow ??$$
And now I'm stuck; any ideas?

Last edited: Aug 18, 2005
2. Aug 18, 2005

### LeonhardEuler

I haven't worked it all out, but it seems like this should lead directly to the solution: Take the equation while it still says
$$\csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right)$$
Now raise both sides to the -1 power:
$$\sin^2\theta = \frac{\cos 2\theta} {2(1+ v_0^{ - 2} gh\sec 2\theta) } = \frac{\cos^2 2\theta} {2\cos 2\theta+ 2v_0^{ - 2} gh}$$
Then substitute for $\sin^2 \theta$:
$$\sin^2\theta = 1 - \cos^2 \theta = 1 - \frac{\cos 2\theta +1} {2}$$
Now all of the trig functions are $\cos 2\theta$ so its just a quadratic equation.

3. Aug 18, 2005

### bomba923

Right...therefore,
$$1 - \cos 2\theta = \frac{{\cos ^2 2\theta }}{{v_0^{ - 2} gh + \cos 2\theta }} \Rightarrow \left( {1 - \cos 2\theta } \right)\left( {v_0^{ - 2} gh + \cos 2\theta } \right) = \cos ^2 2\theta \Rightarrow$$
$$v_0^{ - 2} gh \; + \; \left( {1 - v_0^{ - 2} gh} \right)\cos 2\theta \; - \; 2\cos ^2 2\theta = 0 \Rightarrow 2v_0^2 \cos ^2 2\theta \; - \; \left( {v_0^2 - gh} \right)\cos 2\theta \; - \; gh = 0 \Rightarrow$$

$$\cos 2\theta = \frac{{v_0^2 - gh + \sqrt {v_0^4 + 6v_0^2 gh + g^2 h^2 } }}{{4v_0^2 }} \Rightarrow \theta = \frac{1}{2}\cos ^{ - 1} \left[ {\frac{1}{4} - \frac{{gh - \sqrt {\left( {v_0^2 + gh} \right)^2 + 4v_0^2 gh} }}{{4v_0^2 }}} \right]$$

*Is this correct??
*Can the expression within the arccosine be further simplified?

Last edited: Aug 18, 2005
4. Aug 19, 2005

### Fermat

You get your last expresssion here by dividing by $$cos2\theta$$, but when h = 0, $$Cos2\theta = 0$$.

When projecting upon horizontal ground, the optimum angle is 45º.

In your final expression (last post), what value of $$\theta$$ do you get for h = 0?

5. Aug 19, 2005

### bomba923

Well, the original problem explicitly stated
Somewhat similar as to why $a \ne 0$ when solving $a x ^ 2 + b x + c = 0$ for $x$. You cannot divide by $2 a$ in the quadratic formula if $a = 0$. However, if $a = 0$, then there are simpler ways of solving quadratics and get solutions for x. But why is this an issue here? What's wrong with it, if I consider only $h > 0$ ?

However, we can substitute $h = 10 ^ {-10}$...or any number sufficiently close to zero--->we can also substitute h = 0. Also, let $$v_0 = 10$$. Needless to say, $g = 9.8$.
Therefore, the optimal angle would be:
$$\theta = \frac{1}{2}\cos ^{ - 1} \frac{1}{2} = \frac{\pi }{6}$$

However, the correct answer is $$\theta = {\pi/4}$$.

*Thus, I think my formula for the optimal $$\theta$$ for maximum range is incorrect........But why? Where have I gone wrong? Seriously, why is it wrong do divide by $\cos {2\theta}$...I mean, I did state $h > 0$.
|*It seems there's a problem. But what exactly is it? How can it be solved?
And what is the REAL, the True formula to calculate the $$\theta$$ ?

*According to a friend's calculator, the optimal $$\theta$$ for maximum range is given by the formula:

$$\theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right]$$

But is this truly the formula to find the optimal $$\theta$$ for the maximum range? If so, how is it derived?

Last edited: Aug 19, 2005
6. Aug 19, 2005

### Fermat

In the quadratic formula, you can't have a=0, not only because division by zero is out, but because if a = 0, then you don't have a quadratic, just a linear expression in x. The QF is for quadrtic expressions, so having a != 0 is a condition of using it.

In your optimum angle problem, why do you not allow h=0?
h=0 is the state of the problem when projecting upon horizontal ground and any (general) expression should be able to take h=0 as one possible value.
Also, putting h=0 allows you to check the accuracy of your own general solution for one specific case.

I solved this problem in another forum, some time ago, but I can't find my solution. I'll have another look and see if I can figure out something.

Last edited: Aug 19, 2005
7. Aug 19, 2005

### bomba923

From reading your previous post, it seemed that you saw h = 0 as a problem...a problem when dividing by $cos {2\theta}$ when solving ${dx}/{d\theta} = 0$ for for $$\theta$$. Would it be this step that invalidates my original formula for the $$\theta$$ ?
In some other forum (non-PF)...or just in another section of PF?
Can you remember some phrases/quotes, that I may google-search for it?

Aside from that, how fares the calculator solution?
$$\theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right]$$
Not sure how this was derived...but it appears to work:
@h=0, $\theta = {\pi}/{4}$, and ${d\theta}/{dh} < 0$ as expected.

8. Aug 19, 2005

### Fermat

This can be a problem, but as a problem can be circumvented by taking out $$cos2\theta$$ as a factor, which can be equated it to zero as a solution of the expression. This would give you $$\theta = \pi/4$$ as one solution! So now I don't think that's the problem.
It's from TSR maths forum, but I've already searched for it with no luck. I'm sorry but I can't remember any significant details - but it was the exact same problem. Fiinding the optimum angle for shooting off a cliff/hill top.
Still working on it. But there's a problem in the expression you gave. gh and $$v_0^4$$ should have the same units. Shouldn't it be $$v_0^2$$

9. Aug 19, 2005

### Fermat

Worked throught the problem again, and as answer got,

$$cos2\theta = \frac{gh}{gh + v_0^2}$$

which, in effect, is the same the one you got,

$$\theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right]$$

(once you chage the $$v_0^4$$ to $$v_0^2$$)

Found your error - it was just a wrong sign!

The final term $$+ \; 2v_0^{ - 1} gh$$ should be $$- \; 2v_0^{ - 1} gh$$

If you work through from there on, you should end up with the expression I got, which you can simplify to the one from your friend with the calculator.