- #1

- 759

- 0

I shoot a cannon from a cliff of height [itex] h [/itex], with an initial velocity [itex] v_0 [/itex] and angle of elevation [tex] \theta [/tex]. If [itex] h > 0 [/itex], what [tex] \theta [/tex] 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

[tex] y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2} [/tex]

The cannon will land at ground zero, so I find that the time spent in the air is:

[tex] 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} [/tex]

And therefore the range is:

[tex] 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) [/tex]

*But, which [tex] \theta [/tex] will

[tex] \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 [/tex]

[tex] 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 [/tex]

To simplify this equation, I can multiply both sides by

[tex] \frac{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{{v_0 \sin \theta }} [/tex]

and continue to get :

[tex] \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 [/tex]

[tex] \csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) [/tex]

Now, we can multiply both sides by [itex] \cos 2\theta [/itex], to get

[tex] \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 ?? [/tex]

And now I'm stuck; any ideas?

Ok...

The cannonball's height at any moment within its trajectory is

[tex] y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2} [/tex]

The cannon will land at ground zero, so I find that the time spent in the air is:

[tex] 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} [/tex]

And therefore the range is:

[tex] 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) [/tex]

*But, which [tex] \theta [/tex] will

**maximize**the range? Here, I simply find the roots of [itex] {dx}/{d\theta} [/itex]:[tex] \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 [/tex]

[tex] 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 [/tex]

To simplify this equation, I can multiply both sides by

[tex] \frac{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{{v_0 \sin \theta }} [/tex]

and continue to get :

[tex] \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 [/tex]

[tex] \csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) [/tex]

Now, we can multiply both sides by [itex] \cos 2\theta [/itex], to get

[tex] \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 ?? [/tex]

And now I'm stuck; any ideas?

Last edited: