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Projectile Motion

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant equations

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

    3. 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?
  2. jcsd
  3. Sep 29, 2010 #2


    User Avatar
    Homework Helper

    Hi kvan, welcome to PF.

    Your approach is correct. Show your calculations.
  4. Sep 29, 2010 #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: Sep 30, 2010
  5. Sep 30, 2010 #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?
  6. Sep 30, 2010 #5


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    Homework Helper

    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.
  7. Sep 30, 2010 #6
    I got the answer finally, thanks so much for the help. :D
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