Finding the Launch Angle of an Artillery Gun

AI Thread Summary
To determine the launch angle of an artillery gun firing a shell at 650 m/s to hit a target 12.5 km away, the user initially broke down the velocity into x and y components. They calculated the time to reach maximum height and attempted to use this to find the total time of flight. After several calculations and adjustments using trigonometric identities, they struggled to isolate the angle. Ultimately, they were guided to use the relationship between time of flight and the range to solve for the angle, successfully finding the solution with assistance. The discussion highlights the importance of correctly applying kinematic equations and trigonometric identities in projectile motion problems.
kvan
Messages
8
Reaction score
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 v^2_y = v^2_0_y + 2gt and v_x = \frac{d_x}{t}

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?
 
Physics news on Phys.org
Hi kvan, welcome to PF.

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

y= 650\sin\theta
x= 650\cos\theta

I use the y value and plugged it into

v^2_y= v^2_0_y + 2gt

rearranged into

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

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

\frac {(650\sin\theta\)^2} {9.8}

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

43112(\sin\theta)^2

I then used the equation

v_x = \frac {d_x}{t}

which gives me

650\cos\theta = \frac{12500}{43112(\sin\theta)^2}

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

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

\cos\theta\*(\sin\theta\)^2 = 4.46*10^-4

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

\sin^2\theta = (\sin\theta)^2 = \frac{1-\cos 2\theta}{2}

I replace sin with cos

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

Multiply it through

\frac{\cos\theta - \cos^2 2\theta}{2} = 4.46*10^-^4

multiply both sides by 2

\cos\theta - \cos^2 2\theta =8.92*10^-^4

I factor out cos from the right side

\cos\theta (1-\cos 2\theta ) =8.92*10^-^4

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:
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.

v_f_y = v_f_i - gt

0 = 650\sin\theta - 9.8t

t = \frac{650\sin\theta}{9.8}

T = 2t = 2* \frac{650\sin\theta}{9.8}

T = \frac {650\sin\theta}{4.9}

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

y=0

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

=(\frac{650\sin\theta ^2}{4.9}) - (\frac{650\sin\theta ^2}{4.9})

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?
 
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.
 
I got the answer finally, thanks so much for the help. :D
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top