Projectile motion, need to find angle

In summary, to fire the cannon at a certain angle, you need to solve for tan(angle) and use that to find the angle above the horizontal.
  • #1
frosti
13
0

Homework Statement


A cannon with a muzzle speed of 1009 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 795 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)



Homework Equations


I'm thinking this equation is my best bet, yf = tan(angle)xf - (g/(2 * vi^2 * cos^2(angle))) * (xf^2)


The Attempt at a Solution


since I know the vi, xf, and yf, I was able to plug everything in and reached 795 = 1900tan(angle) - 17.375/cos^2(angle). This is where I am stuck. I tried several times but could not find a way to solve for angle.

I would really appreciate it if someone could point out how to solve for the angle in my problem. Or is there a totally different way of solving this problem?
 
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  • #2
Your work DOES look great (I checked your first equation). The problem is that you've got an algebraically difficult thing to solve for (the angle) because it's inside trig functions.

What I suggest is keeping your equation in terms of the components. Relate one of these components ONLY to the initial speed and the remaining component. Then solve for the component you kept. You can use this solution to find the angle.

Added at a later edit: You could also just graph your second function and look for where it crosses zero, but that's easy only because of modern graphing tools.
 
Last edited:
  • #3
thank you for your input, i separated my problem into x and y components:
x component ---> 1900 = 1009cos(theta) * t
y component ---> 795 = 1009sin(theta) * t - 4.9t^2

I used my x component to solve for cosine
1900/1009 = cos(theta) * t
cos(theta) = 1.883/t

I then used trig identity sin^2(theta) + cos^2(theta) = 1
sin^2(theta) + (1.883/t)^2 = 1
sin(theta) = square root(1 - (1.883/t)^2)

then I plugged sin(theta) equation into my y component
795 = 1009 * square root(1-(1.883/t)^2) * t -4.9t^2

I think I'm on the right track but I don't really know where to go with this, it's not really a quadratic equation and I don't know what to do now. Any pointer would be greatly appreciated.
 
  • #4
nm, I got it. Thank you for all your help.
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air under the force of gravity. It is a combination of horizontal and vertical motion.

2. How do you find the angle in projectile motion?

The angle in projectile motion can be found using trigonometric functions. The angle can be calculated by taking the inverse tangent of the vertical velocity divided by the horizontal velocity.

3. What factors affect the angle in projectile motion?

The angle in projectile motion is affected by the initial velocity, the acceleration due to gravity, and the trajectory of the object. The angle can also be affected by air resistance or other external forces.

4. How does the angle affect the range of a projectile?

The angle in projectile motion affects the range of the object. The optimal angle for maximum range is 45 degrees. A lower angle will result in a shorter range, while a higher angle will result in a higher range.

5. Can the angle in projectile motion be negative?

Yes, the angle in projectile motion can be negative. In this case, the object is moving in the opposite direction of the positive x-axis. The angle would be measured in the clockwise direction from the negative x-axis.

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