Projectile - what angle?

jeepey · Feb 27, 2007, 07:41 AM

Quote:

I'm bad at this part of math.

your not the only one! Lol

Capuchin · Feb 27, 2007, 07:46 AM

Hopefully asterisk will pop his head in, I told him to, he'll hopefully have some time to help right now

asterisk_man · Feb 27, 2007, 08:21 AM

Somewhat busy right now so don't hold me to this!

Ok. I follow to
$0.7 = \frac{19sin\theta}{cos\theta} - 9.025cos^2\theta$
The real problem is solving for $\theta$ ...obviously
Sin/cos=tan and
$\cos^2 \theta = \frac {1 + \cos (2\theta ) } 2 \\<br /> \text {so} \\<br /> 0.7=19 \tan \theta - 4.5125 - 4.5125 \cos (2\theta) \\<br /> 5.2125= 19 \tan \theta - 4.5125 \cos (2\theta) <br />$

Unfortunately my brain is unable to figure out how you solve for theta at the moment
Numerically I think the answer is:
0.41113094848254 radians = 23.5560682 degrees

If I substitute into the first equation I listed then I do get 0.7

Someone else can try to substitute this back into the original equations to see if it really does provide the correct answer. Or i'll try more later if I get a chance.

Sorry I couldn't provide more help. I also dislike trig identities

jeepey · Feb 27, 2007, 10:46 PM

Thanks for the help guys, this problem is more complicated than 1st seems. Some parts of the workings I am struggling to follow, but have helped.

jeepey · Mar 5, 2007, 05:24 AM

Sorry I am re visiting this question

I'm not sure if I'm right, but would the -9.025cos^2 theta not be -9.025cos^-2 theta as its being devided by?

Capuchin · Mar 5, 2007, 05:30 AM

I don't think it is being divided by..

Projectile - what angle?

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