Calculating angle for trajectory with exit position not at 0,0

In summary, the conversation discusses using a cannon to shoot a projectile at a target, with the projectile exiting from a point other than 0, 0. The equations for calculating the angle of the cannon and solving for theta using the Quadratic Formula are mentioned.
  • #1
xal
1
0
I'd like to calculate the angle to shoot a projectile from a cannon in order to hit a target. However, the position that the projectile will be exiting from will not be at 0, 0 -- it will vary according to the angle of the cannon itself. The projectile will be exiting from the tip of the cannon, but the cannon will be rotated about a different point (both points, as well as the target position, will be on the same plane though).

If the projectile was exiting from 0, 0, then I could use this:
[tex]\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\mbox {{\tt `\&+-`}
} \left( \sqrt {{{\it v0}}^{4}-{g}^{2}{x}^{2}-2\,gy{{\it v0}}^{2}}
\right) }{gx}}
[/tex]

However, in this case, I was considering replacing [tex]x[/tex] and [tex]y[/tex] with [tex]x-l\cos \left( \theta \right)[/tex] and [tex]y-l\sin \left( \theta \right) [/tex] respectively, where [tex]l[/tex] would be the distance away the tip.

That results in this:
[tex]\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\sqrt {{{\it v0}}^{
4}-g \left( g \left( x-l\cos \left( \theta \right) \right) ^{2}+2\,
\left( y-l\sin \left( \theta \right) \right) {{\it v0}}^{2} \right)
}}{g \left( x-l\cos \left( \theta \right) \right) }}


[/tex]

Different form:
[tex]-1/2\,{\frac {g \left( x-l\cos \left( \theta \right) \right) ^{2}
\left( \tan \left( \theta \right) \right) ^{2}}{{v}^{2}}}+x\tan
\left( \theta \right) -1/2\,{\frac {g \left( x-l\cos \left( \theta
\right) \right) ^{2}}{{v}^{2}}}-y+l\sin \left( \theta \right) =0
[/tex]

But I am not sure if this is a way to do it, and I am not sure to go about solving for the angle either.
 
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  • #2
Any help would be greatly appreciated! The above equation can be solved for theta by using the Quadratic Formula. Specifically, you can rearrange the equation to the form ax^2 + bx + c = 0 and then solve for x. Once you have x, you can take its inverse tangent to find the angle theta.
 
  • #3


I would first suggest verifying the equations and assumptions used in the calculations. It is important to ensure that the equations accurately represent the physical system being studied.

Assuming the equations are correct, the next step would be to solve for the angle using mathematical methods such as substitution or graphical methods. It may also be helpful to plot the trajectory of the projectile and the position of the cannon to visually understand the problem and potential solutions.

Additionally, it may be useful to consider the effects of air resistance and other external factors on the trajectory of the projectile. These factors could affect the accuracy of the calculated angle and should be taken into account in the analysis.

Overall, the key is to approach the problem systematically and accurately, using appropriate equations and methods to solve for the desired angle.
 

1. How do you calculate the angle for a trajectory when the exit position is not at (0,0)?

To calculate the angle for a trajectory with an exit position not at (0,0), you will need to use the trigonometric functions sine, cosine, and tangent. First, determine the horizontal and vertical distance between the starting point and the exit position. Then, use the tangent function to find the angle of elevation or depression.

2. What is the formula for calculating the angle of elevation or depression?

The formula for calculating the angle of elevation or depression is tan(theta) = opposite/adjacent, where theta represents the angle, opposite is the vertical distance, and adjacent is the horizontal distance between the starting point and the exit position.

3. Can you provide an example of calculating the angle for a trajectory with an exit position not at (0,0)?

Yes, for example, if a projectile is launched from a point (2,3) and lands at a point (6,9), the horizontal distance is 4 units and the vertical distance is 6 units. Using the formula, we can calculate the angle of elevation as tan(theta) = 6/4 = 1.5. Using a calculator, we can find that the angle is approximately 56.31 degrees.

4. How does the angle of elevation affect the trajectory of a projectile?

The angle of elevation affects the trajectory of a projectile by determining its vertical and horizontal components. A higher angle of elevation will result in a higher vertical component, causing the projectile to travel higher and farther before hitting the ground. A lower angle of elevation will result in a lower vertical component, causing the projectile to travel a shorter distance before hitting the ground.

5. Are there any other factors that influence the angle of trajectory with an exit position not at (0,0)?

Yes, there are a few other factors that can influence the angle of trajectory with an exit position not at (0,0). These include air resistance, wind speed and direction, and the initial velocity of the projectile. These factors can all impact the trajectory of the projectile and may require adjustments to the calculated angle for more accurate results.

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