# Bullet trajectory; solving for required angle of elevation

1. May 26, 2010

### TylerJFisher

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