A projectile is fired at v0 = 394.0 m/s at an angle of θ = 68.7o with respect to the horizontal. Assume that air friction will shorten the range by 28.1%. How far will the projectile travel in the horizontal direction, R?
To find the horizontal distance traveled by the projectile, we need to calculate the range (R). The range can be determined using the horizontal component of the initial velocity (v0x) and the time of flight (T).
1. Calculate the horizontal component of the initial velocity (v0x):
v0x = v0 * cos(θ)
where v0 is the initial velocity (394.0 m/s) and θ is the angle of projection (68.7 degrees).
v0x = 394.0 * cos(68.7)
v0x ≈ 394.0 * 0.317
v0x ≈ 125.198 m/s
2. Calculate the time of flight (T):
The time of flight is the total time the projectile is in the air before it hits the ground. We can calculate it using the vertical component of the initial velocity (v0y) and the acceleration due to gravity (g).
v0y = v0 * sin(θ)
where v0 is the initial velocity (394.0 m/s) and θ is the angle of projection (68.7 degrees).
v0y = 394.0 * sin(68.7)
v0y ≈ 394.0 * 0.945
v0y ≈ 372.63 m/s
T = 2 * v0y / g
where g is the acceleration due to gravity (9.8 m/s^2).
T = 2 * 372.63 / 9.8
T ≈ 76.021 seconds
3. Calculate the range (R):
The range is the horizontal distance traveled by the projectile, which can be calculated using the formula:
R = v0x * T
where v0x is the horizontal component of the initial velocity (125.198 m/s) and T is the time of flight (76.021 seconds).
R = 125.198 * 76.021
R ≈ 9506.61 meters
Therefore, the projectile will travel approximately 9506.61 meters in the horizontal direction.