Finding the Launch Angle of an Artillery Gun

In summary, if the gun is fired at 650 ms-1 at an angle of 45 degrees above the horizontal, the gun will hit the target.
  • #1
kvan
9
0

Homework Statement



An artillery gun is trying to hit a target that is 12.5 km away. If the shell is fired at 650 ms-1 at what angle above the horizontal should the gun be fired to hit the target assuming that air resistance is negligible? [g=9.8 ms-2]

Homework Equations



I tried using [tex] v^2_y = v^2_0_y + 2gt [/tex] and [tex] v_x = \frac{d_x}{t} [/tex]

The Attempt at a Solution



I tried breaking the 650m/s into component vectors for x and y. I used the y component to try and find time it took to reach maximum height, then doubled this value to find total time. I used this value for time and plugged it into v=d/t in an attempt to maybe create an equation where I could find the angle for the cannon.

I got stuck at that point and I don't really know what to do. Any suggestions?
 
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  • #2
Hi kvan, welcome to PF.

Your approach is correct. Show your calculations.
 
  • #3
Thanks, well my component vectors look like this

y= [tex] 650\sin\theta [/tex]
x= [tex] 650\cos\theta [/tex]

I use the y value and plugged it into

[tex] v^2_y= v^2_0_y + 2gt [/tex]

rearranged into

[tex] \frac {t = v^2_y - v^2_0_y}{2g} [/tex]
[tex] \frac {0^2 - (650\sin\theta)^2}{-9.8*2} [/tex]

since this value for t is just the time taken to reach maximum height I doubled it giving me

[tex] \frac {(650\sin\theta\)^2} {9.8}[/tex]

I tried to simplify the value so I squared the 650 and divided by 9.8

[tex] 43112(\sin\theta)^2 [/tex]

I then used the equation

[tex] v_x = \frac {d_x}{t} [/tex]

which gives me

[tex] 650\cos\theta = \frac{12500}{43112(\sin\theta)^2}[/tex]

I had converted the 12.5km to 12500m. I attempt to move all the values of [tex] \theta [/tex] to one side

[tex] \cos\theta\ = \frac{4.46*10^-4}{(\sin\theta)^2} [/tex]

[tex] \cos\theta\*(\sin\theta\)^2 = 4.46*10^-4 [/tex]

I used trig identities to try and simplify the right side to only 1 trig function

[tex] \sin^2\theta = (\sin\theta)^2 = \frac{1-\cos 2\theta}{2} [/tex]

I replace sin with cos

[tex] \cos\theta(\frac{1-\cos 2\theta}{2})=4.46*10^-^4[/tex]

Multiply it through

[tex] \frac{\cos\theta - \cos^2 2\theta}{2} = 4.46*10^-^4[/tex]

multiply both sides by 2

[tex] \cos\theta - \cos^2 2\theta =8.92*10^-^4 [/tex]

I factor out cos from the right side

[tex] \cos\theta (1-\cos 2\theta ) =8.92*10^-^4 [/tex]

That is basically all of my calculations. I get stuck at this point, any tips or pointers you guys can give?
Sorry if the math equations are hard to read, I'm kinda figuring out latex as I go along
 
Last edited:
  • #4
Oh man, I can't believe I messed that up. I tried using the method you put but I don't think it works, or I did something wrong.

[tex] v_f_y = v_f_i - gt [/tex]

[tex] 0 = 650\sin\theta - 9.8t[/tex]

[tex] t = \frac{650\sin\theta}{9.8} [/tex]

[tex] T = 2t = 2* \frac{650\sin\theta}{9.8}[/tex]

[tex] T = \frac {650\sin\theta}{4.9}[/tex]

[tex] y=v_o\sin\theta*t-\frac{1}{2}gT^2[/tex]

[tex] y=0[/tex]

[tex] 0=650\sin\theta*\frac{650\sin\theta}{4.9}-4.9*(\frac{650\sin\theta}{4.9})^2[/tex]

[tex] =(\frac{650\sin\theta ^2}{4.9}) - (\frac{650\sin\theta ^2}{4.9}) [/tex]

I'm not really sure how to solve for the angle at this point since both sides equal each other. But did I make a mistake in my calculations?
 
  • #5
In the problem range is given and the velocity is given. Then

Time of flight T = 12500m/650*cosθ.

Now y = Vo*sinθ*Τ - 1/2*g*T^2.

Put y = 0 and substitute the value of T and solve for θ.
 
  • #6
I got the answer finally, thanks so much for the help. :D
 

1. How is the launch angle of an artillery gun determined?

The launch angle of an artillery gun is determined by using mathematical equations and calculations based on the gun's muzzle velocity, projectile weight, and desired range. The goal is to find the angle at which the projectile will travel the farthest distance.

2. Why is it important to find the launch angle of an artillery gun?

The launch angle of an artillery gun is crucial in determining the trajectory and range of the projectile. It also helps in accurately aiming and hitting the intended target.

3. What factors can affect the launch angle of an artillery gun?

The launch angle of an artillery gun can be affected by various factors such as air resistance, wind speed and direction, temperature, altitude, and the type and condition of the gun and ammunition.

4. How does the launch angle of an artillery gun impact its accuracy?

The launch angle of an artillery gun greatly affects its accuracy. A wrong or miscalculated angle can result in the projectile falling short or overshooting the target. Therefore, finding the correct launch angle is crucial for achieving the desired accuracy.

5. Can the launch angle of an artillery gun be adjusted?

Yes, the launch angle of an artillery gun can be adjusted by changing the elevation of the gun or by using additional tools such as a quadrant or clinometer. This allows for more precise and accurate targeting based on the conditions and range of the intended target.

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