Maximizing Cannon Range from a Cliff of Height h

[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?

 

[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.

 

[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?

 

[Fermat]

...
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


...

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?

 

[bomba923]

Well, the original problem explicitly stated

If h > 0, what \theta .. ..

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?

 

[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.

 

[bomba923]

In your optimum angle problem, why do you not allow h=0?

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 ?

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.

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.

 

[Fermat]

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 ?

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.

... 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?

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.

...
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.

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

 

[Fermat]

Worked through 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!

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

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.