Firing Cannon to Start Avalanche: Angle Calculation

[deezy]

Homework Statement

A cannon has a muzzle speed of 1000 m/s, and it is used to start an avalanche on a mountain 1000 m from the cannon horizontally and 500 m above the cannon. What angle above the horizontal should the cannon be fired?

$$v_0 = 1000 m/s$$
$$y_t = 500 m$$
$$x_t = 1000 m$$

Homework Equations

$$x_t = x_0 + v_0 cos \theta t$$
$$y_t = y_0 + v_0 sin \theta t - g t^2$$

The Attempt at a Solution

I don't know if this approach is right but I tried a system of equations:

$$1000 = 1000 cos \theta t$$
$$1 = cos \theta t$$
$$t = \frac{1}{cos \theta}$$

I tried substituting t from the first equation into this one:
$$500 = 1000 sin \theta t - (1/2)(9.81)t^2$$
$$500 = 1000 sin \theta t - (1/2)(9.81)t^2$$
$$500 = 1000 \frac {sin \theta }{cos \theta} - \frac{(4.905)}{cos ^2 \theta}$$
$$500 = 1000 tan \theta - (4.905)sec \theta$$

But I didn't know how to solve for theta. Is this the right approach, or is there an easier way to do this problem?

[tex][/tex]

 

[vela]

You should get in the habit of plugging numbers in only near the end. Using a trig identity, you get:
\begin{align*}
y_t &= x_t \tan\theta - x_t^2\left(\frac{g}{2v_0^2}\right)\sec^2 \theta \\
&= x_t \tan\theta - x_t^2\left(\frac{g}{2v_0^2}\right)(1+\tan^2 \theta)
\end{align*}
That's quadratic in tan θ. (I'd plug the numbers in now.) Can you take it from here?

 

[deezy]

Plugging in from that point I got:

$$500 = 1000 tan \theta - \frac {1000^2}{2*1000^2}*9.81(1+tan^2 \theta)$$

Which simplified to:

$$tan^2 \theta - 203.874 tan \theta + 102.937 = 0$$

I get stuck here. Not sure how to solve this... my idea was let $$x = tan \theta$$ and solve x using the quadratic formula, then plug $$tan \theta$$ back in for x to find the angle. Does this work?
[tex][/tex]

 

[vela]

Yes, that'll work.