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

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Homework Help Overview

The problem involves determining the angle at which a cannon should be fired to hit a target located at a specific horizontal and vertical distance. The context is projectile motion, specifically applying the quadratic equation to find the angle of launch given certain parameters.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up the equations of motion for the cannonball, leading to a quadratic equation in terms of the tangent of the angle. There are attempts to manipulate the equations to isolate the angle, with some participants expressing confusion about the next steps.

Discussion Status

Several participants have provided intermediate results and expressed uncertainty about the solutions obtained from the quadratic equation. There is an ongoing exploration of the implications of the results, particularly regarding the validity of the angles derived from the quadratic solutions.

Contextual Notes

Participants emphasize the importance of precision in calculations, noting that rounding off values can lead to significant errors in the context of trigonometric functions. There is also a mention of the sensitivity of the tangent function to small changes in its argument.

neutron star
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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 :\.
 
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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.
 


I got x=-15.37 x=-3984.63
 


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.
 

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