What Angle Should a Cannon Be Fired to Hit a Target 2000m Away and 800m Above?

[tuna_wasabi]

I'll post these in two messages to make this easier. First problem:

A cannon with a muzzle speed of 1000 m/s is fired at a target 2000m away horizontally and 800m above the cannon vertically. Find the necessary angle to hit the target.

I'm pretty stuck on this one. I've tried juggling all the kinematics equations but the best I can do is solve x=v*cos(theta)*t for t and drop it into y=(1/2)gt^2 + v*sin(theta)*t. This yields the standard Range equation, but I can't solve that for theta. I'm just not sure how to derive the right equation.

 

[tuna_wasabi]

Never mind the second problem, I figured it out as I was typing it. Thanks in advance.

 

[dimensionless]

First off, gravity counters the vertical motion of the cannon ball. Your y-equation should thus be y(t)=-(1/2)gt^2+v*sin(theta)*t

To find theta you will need to determine how long it will take the ball to x(t)=2000m. Once you have found 't' you can then substitute it into your y-equation and solve for theta.

 

[tuna_wasabi]

Yes, but to know how long it takes the ball to travel 2000m along the ground I need to know the x-component of the initial velocity. But I can't get the x-component without theta, which is what I'm trying to find.

 

[finchie_88]

You could always combine the x and y component equations (eliminating t in the process), and solve the resulting equation for the angle, it would probably involve more work, but it would work best overall.

 

[tuna_wasabi]

As stated above, I've tried that. Solving x=vi*cos(theta)t for t and putting into y=(1/2)gt^2 + v*sin(theta)*t yields the standard projectile motion range equation, which is [deep breath]:

y = tan(theta)*x - [(g*x^2)/(2*vi^2*cos^2(theta))]

If you can solve that for theta, I will bow to your awesome trig skills.

 

[siddharth]

Try using

<br /> \frac{1}{\cos^2 \theta} = sec^2 \theta

sec^2\theta = 1 + \tan^2 \theta

and solve for \tan \theta.