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Bullet trajectory; solving for required angle of elevation

  1. May 26, 2010 #1
    The problem statement, all variables and given/known data
    Bob is trying to hit a stationary target 146.00m downrange, his eyes are 1.8m above the ground, firing from a standing position. The muzzle velocity of the projectile is 382.524m/sec (1255 ft/sec), wind is negligible; at what firing angle must he fire to successfully hit the target? The target is 1m above the horizontal.

    DATA
    Variables
    T: projectile flight time in seconds
    V: projectile velocity (m/s)
    g: gravity in m/s^2
    a (theta): required firing angle in degrees

    Test/develop environment: Controlled
    Overview: Set up markers to a distance of 200m, 1 marker per 25m
    Program will not compensate for cross-winds, humidity, etc.

    Required information:
    1)Drag factor
    2)Launch angle
    3)Initial velocity
    4)Gravity (Default: 9,8022m/s^2)
    5)Shape of parabola

    Variables:
    T: Projectile flight time in seconds
    V: Projectile velocity in m/s
    g: gravity in m/s^2
    a: Required firing angle in degrees

    Test-caliber: Winchester Wildcat .22 High Velocity
    Point: Solid
    Bullet weight: 40 grains (2.59196 grams)
    Muzzle velocity (m/s): 382.524
    Velocity @ 100m: 1017 (18.96% decrease
    Muzzle energy (ft/lbs):139.82
    Energy @ 100m (ft/lbs):91.82 (34.33% decrease)
    Ballistic coefficient: .100

    Attempt at a solution:
    Would I use this formula to find the angle of elevation?

    "Angle θ required to hit coordinate (x,y)" @ http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

    θ = tan-1((v2 +/- ((v4-g(gx2+2yv2)))0.5)/gx)

    The above formula does not compensate for air resistance, so should I scrap that formula and seek an alternative derivation?
     
    Last edited: May 26, 2010
  2. jcsd
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