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Formula for the angle a sniper must make to hit a target at distance x

  1. Feb 26, 2012 #1
    I just got curious today and tried to determine the angle a sniper must make with his barrel to hit a target over some distance, neglecting air resistance, humidity, etc. Any suggestions as to why my solution on the bottom right is wrong?
     

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  2. jcsd
  3. Feb 26, 2012 #2

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    You have y=yo+vo t - g t^2/2 and then you have 0 = vo t - g t^2/2 so that means y=yo always, and thats not what you want.
     
  4. Feb 26, 2012 #3
    Your handwriting is neat but your image is on it's side :(

    The standard equations of motion for an object in a uniform gravitational field is

    [itex]y=y_0 + v_y_0 t -\frac{1}{2} g\ t^2[/itex]
    [itex]x=x_0+v_x_0 t[/itex]

    Edit;
    LaTeX doesn't work on these boards?
     
  5. Feb 26, 2012 #4
    Yes, I did this to solve for T. Afterwords I multiplied that formula by 2 in order to determine the total air time. Once I found the total air time formula T= 2Fsin(θ)/2 I plugged it into the formula X = X° + Vxo * T. Was this an incorrect strategy?
     
  6. Feb 27, 2012 #5

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    But if you say that y=y0, thats wrong. You are saying that the y coordinate never changes, always stays where it started from. If you make that wrong assumption, you are bound to get wrong results.


    You wrote it wrong, a_b_c is not allowed, maybe a_{b_c} or {a_b}_c
     
  7. Feb 27, 2012 #6
    He substituded 0 for y_0, because y_0 happens to be 0, and then set the remaining expression for y equal to 0. There's nothing wrong with that

    However, there's an error (or even 2 errors) between

    [tex] x = \frac { 2 F \cos {\theta})F \sin {\theta} } {g} [/tex]

    and

    [tex] x = \frac {F \sin {2 \theta} } {g} [/tex]
     
  8. Feb 27, 2012 #7
    How might you have it willem2?
     
  9. Feb 27, 2012 #8
    I get

    [tex] x = \frac {F^2 sin {2 \theta}} {g} [/tex]
     
  10. Feb 27, 2012 #9
    Sooo the formula for the angle needed is? I don't mean for this to sound rude.

    I solved another problem in my text book very similar to this one where a fire hose shot water at a velocity of 6.5 m/s. I had to determine was angle(s) the nozzle could be at to reach a target 2.5 meters away. I worked the problem very similarly to my OP and found the angle of 18* to work. The book noted a second angle at 72*. I suppose 90 - 18 = 72 so I can somewhat visualize this second result though still fail and finding it personally. The equation I used was the same for my sniper question...

    (1/2)arcsin(xg/v^2) = theta

    Where x is displacement in meters
    G is the gravitational constant in m/s^2
    V is the velocity in m/s

    I also did a dimensional analysis on the argument of arcsin and found all the units cancelled perfectly. Could this truely a working formula?
     
  11. Feb 27, 2012 #10

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    Ok, I see, the t is the t when the bullet hits the aim point.
     
  12. Feb 27, 2012 #11
    lol just used Firefox as my browser and willem2s formula looks far more comprehendable

    Yes willem2 I did reach that formula after some revisions. I didn't square the Velocity of the projectile (bullet). Although after solving the same equation for theta are you able to reach my formula of

    (1/2)(arcsin(XG/V^2) = theta

    ? Idk the fancy coding needed to make this formula look neat yet hah
     
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