So, what, I just take the first one?Yes, take the first one.

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 5K views
neutron star
Messages
78
Reaction score
1
Now what? Quadratic equation!

Homework Statement


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.


Homework Equations





The Attempt at a Solution


Xf=Xo+Vox t = Xo+Vo cosΘt
Yf=Yo+Voy t -1/2gt[tex]^2[/tex] = -Yo+VosinΘt-1/2gt[tex]^2[/tex]

t=2000/1000cosΘ = 2/cosΘ

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

800=tanΘ-2g(1+tan[tex]^2[/tex]Θ)

ax[tex]^2[/tex]+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 :\.
 
Physics news on Phys.org


Set [tex]\tan{\theta}\equiv Z[/tex]

You know have a quadratic equation in [tex]Z[/tex]
 


19.62z[tex]^2[/tex]+2000z-780.38

or

20z[tex]^2[/tex]+2000z-780

But now what?
 


neutron star said:
19.62z[tex]^2[/tex]+2000z-780.38

or

20z[tex]^2[/tex]+2000z-780

But now what?

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

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


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

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

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

I did that and got weird answers.
 


neutron star said:
I got x=-15.37 x=-3984.63

You plugged your numbers in wrong.

I got:

[tex]Z_1\approx 0.3887[/tex]

This correlates to: [tex]\theta=21.241^o[/tex]

[tex]Z_2\approx-102.3255[/tex]

This correlates to: [tex]\theta=-89.44^o[/tex] which is utter nonsense.