1. Sep 29, 2009

neutron star

1. The problem statement, all variables and given/known data
A cannon having a muzzle speed of 1000 m/s is used to destroy a target on a mountaintop. The target is 2000m from the cannon horizontally and 800m above the ground. At what angle, relative to the ground, should the cannon be fired? Ignore air friction.

2. Relevant equations

3. The attempt at a solution
Xf=Xo+Vox t = Xo+Vo cosΘt
Yf=Yo+Voy t -1/2gt$$^2$$ = -Yo+VosinΘt-1/2gt$$^2$$

t=2000/1000cosΘ = 2/cosΘ

800=1000sinΘ(2/cosΘ) - 1/2g (2/cosΘ)$$^2$$
=2000tanΘ-1/2g 4/cos$$^2$$Θ
1/cos$$^2$$-sec$$^2$$=1+tan$$^2$$

800=tanΘ-2g(1+tan$$^2$$Θ)

ax$$^2$$+bx+c=0

Ok, what do I do to get the angle now, grr I'm drawing a blank, I don't have long to finish this :\.

2. Sep 29, 2009

RoyalCat

Set $$\tan{\theta}\equiv Z$$

You know have a quadratic equation in $$Z$$

3. Sep 29, 2009

neutron star

19.62z$$^2$$+2000z-780.38

or

20z$$^2$$+2000z-780

But now what?

4. Sep 29, 2009

RoyalCat

Don't round those off. That's a very bad idea, especially since you're going to deal with $$\tan{\theta}$$ soon and that function is sensitive to such small changes.

You have a quadratic equation, how do you solve a quadratic equation?

5. Sep 29, 2009

neutron star

-b+or- sq root b^2-4ac all over 2a

I did that and got weird answers.

6. Sep 29, 2009

neutron star

I got x=-15.37 x=-3984.63

7. Sep 29, 2009

RoyalCat

You plugged your numbers in wrong.

I got:

$$Z_1\approx 0.3887$$

This correlates to: $$\theta=21.241^o$$

$$Z_2\approx-102.3255$$

This correlates to: $$\theta=-89.44^o$$ which is utter nonsense.
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