- #1
neutron star
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Now what? Quadratic equation!
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.
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 :\.
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 :\.