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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}((v^{2}+/- ((v^{4}-g(gx^{2}+2yv^{2})))^{0.5})/gx)

The above formula does not compensate for air resistance, so should I scrap that formula and seek an alternative derivation?

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# Bullet trajectory; solving for required angle of elevation

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