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?

[tex]v_0 = 1000 m/s[/tex]
[tex]y_t = 500 m[/tex]
[tex]x_t = 1000 m[/tex]

Homework Equations

[tex]x_t = x_0 + v_0 cos \theta t[/tex]
[tex]y_t = y_0 + v_0 sin \theta t - g t^2[/tex]

The Attempt at a Solution

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

[tex]1000 = 1000 cos \theta t[/tex]
[tex]1 = cos \theta t[/tex]
[tex]t = \frac{1}{cos \theta} [/tex]

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

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:

[tex]500 = 1000 tan \theta - \frac {1000^2}{2*1000^2}*9.81(1+tan^2 \theta)[/tex]

Which simplified to:

[tex] tan^2 \theta - 203.874 tan \theta + 102.937 = 0 [/tex]

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

 

[vela]

Yes, that'll work.

 

[rwisz]

I would approach this problem by first considering the relevant physical principles at play. In this case, we are dealing with projectile motion and the effects of gravity. The goal is to find the angle at which the cannon should be fired in order to achieve the desired horizontal and vertical distances.

To start, I would draw a diagram of the situation, with the cannon at the origin (x=0, y=0) and the avalanche starting at (x=1000, y=500). The initial velocity of the cannonball is 1000 m/s, and we want to find the angle at which it should be fired.

Next, I would use the equations of motion for projectile motion to set up a system of equations. We know that the horizontal distance traveled (x_t) is equal to the initial horizontal velocity (v_0x) multiplied by the time (t), which is equal to the initial velocity (v_0) multiplied by the cosine of the angle (\theta). Similarly, the vertical distance traveled (y_t) is equal to the initial vertical velocity (v_0y) multiplied by the time (t), minus half the acceleration due to gravity (g) multiplied by the square of the time (t^2). We also know that the initial horizontal velocity is equal to the initial velocity multiplied by the cosine of the angle, and the initial vertical velocity is equal to the initial velocity multiplied by the sine of the angle.

Therefore, our system of equations would be:

x_t = v_0 cos \theta t
y_t = v_0 sin \theta t - \frac{1}{2}gt^2
v_0x = v_0 cos \theta
v_0y = v_0 sin \theta

Solving for t in the first equation, we get:

t = \frac{x_t}{v_0 cos \theta}

Substituting this into the second equation, we get:

y_t = v_0 sin \theta \frac{x_t}{v_0 cos \theta} - \frac{1}{2}g \left(\frac{x_t}{v_0 cos \theta}\right)^2

Simplifying, we get:

y_t = x_t tan \theta - \frac{g}{2v_0^2} x_t^2 sec^2 \theta

Now, we can substitute in the given values for x_t,