Finding the Launch Angle of an Artillery Gun

[kvan]

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?

 

[rl.bhat]

Hi kvan, welcome to PF.

Your approach is correct. Show your calculations.

 

[kvan]

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

 

[kvan]

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?

 

[rl.bhat]

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.

 

[kvan]

I got the answer finally, thanks so much for the help. :D