Solving for Angle in Projectile Motion

[blackboy]

Homework Statement

A cannon with muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. The target 2000m from the cannon horizontally and 800m above the cannon. At what angle should the cannon be fired?

Homework Equations

A buttload of the projectile motion ones.

The Attempt at a Solution

Ok what I did was I tried to solve for t and get equations for the angle, but my lack of trig gets in the way. Ok for no better thing to use let =theta. I get 1000cos()t=2000 so cos()t=2. Then for the y component I get 800=sin()t-4.9t^2. I solved here for t with the quadratic equation. Substituting the resulting gives me 19.6=500sin(2)+64sqrt[15625sin^2()-245]. I don't know how to solve for . Help please! Thank you!

 

[dx]

Then for the y component I get 800=sin()t-4.9t^2.

Should be 800 = 1000sin()t - 4.9t².

 

[blackboy]

Yeah I just forgot to type that there, however I do not know how to solve the trig equation there.

 

[Pengwuino]

Using your kinematic equations, you know...

$$\begin{array}{l}<br /> y = \sin (\theta )v_0 t + \frac{{gt^2 }}{2} \\ <br /> x = \cos (\theta )v_0 t \\ <br /> \end{array}$$

If you isolate the trigonometric parts on the right side, you can divide the equations and find $$\frac{{y - \frac{{gt^2 }}{2}}}{x} = \frac{{\sin (\theta )v_0 }}{{\cos (\theta )v_0 }}$$

Now the problem is that t is still in there. I would suggest substituting $$t = \frac{x}{{\cos (\theta )v_0 }}$$. Do some manipulation and you can determine your angle! If you need to use trigonometric identities, i highly suggest checking out http://en.wikipedia.org/wiki/Trigonometric_identities and really getting down the use of the identities.

Note, there are 2 solutions to this problem. This is simply because there is a very large angle that you can shoot the cannon upwards such that it'll hit its target after its long but narrow arc is completed and it's falling downward.

 

[dx]

Suppressing the constants, you should get an equation like

$$\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos^2 \theta} = 1$$

Wrtie sin(θ)/cos(θ) as tan(θ), and 1/cos²(θ) as 1 + tan²(θ).

Putting constants back, you will get a quadratic in tan like tan²(θ) + B tan(θ) + C = 0.

 

[blackboy]

Solving for we get tan()=0,-1. That is an angle of 0 degrees or 135 or 225? What?

Oops sorry forgot about the constants..

 

[dx]

What equation did you solve? This is the one you should have gotten:

819.6 = 2000 tan(θ) - 19.6 tan²(θ)

 

[blackboy]

Um can you show me where you got those constants from?

 

[dx]

You had the equation

800 = 1000sin(θ)T - 4.9T²

Now just substitue T = 2/cos(θ) in this.

 

[blackboy]

Ok I learned that 1/cos^2()= Tan^2()+1. That helped a lot so solving for , I finally get =22.36 or 89.43. Thanks a bunch both of you! BTW when did you guys learn these trig identities? I did not see that 1/cos^2()=tan^2()+1 on the wikipedia page.

 

[dx]

I didn't see it at first either. I was breaking my head trying to solve it for cos(θ)! I am surprised 1/cos^2(θ)=tan^2(θ)+1 is not listed on wiki, it's a pretty standard identity, usually written in the form sec^2(θ) = 1 + tan^2(θ).

 

[Pengwuino]

It is listed on the site in the section http://en.wikipedia.org/wiki/Trigonometric_identities#Basic_relationships . Just got to do the squaring by yourself.