Bullet trajectory; solving for required angle of elevation

[TylerJFisher]

Homework Statement
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?

 

[kuruman]

I think you can safely ignore air resistance, the test caliber and all that useless information below it because none of that is given nor is its need implied in the statement of the problem. Just use the formula you quoted.