Final answer:
The bullet fired at a 60° angle with an initial velocity of 200.0 m/s has a time of flight of approximately 35.35 seconds and reaches a maximum height of about 1532.87 meters.
Step-by-step explanation:
A bullet is fired at an angle of 60° with an initial velocity of 200.0 m/s, and we are interested in determining how long the bullet is in the air (time of flight) and the maximum height it reaches.
To calculate the time of flight, we first determine the vertical component of the initial velocity (Vy) using the equation Vy = V * sin(θ), where V is the initial velocity and θ is the angle of projection. Since the angle given is 60°, Vy = 200 m/s * sin(60°) = 173.2 m/s. The time of flight (T) is then calculated using the kinematic equation for projectiles, T = 2 * Vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s²). Thus, T = 2 * 173.2 m/s / 9.8 m/s² ≈ 35.35 s.
To find the maximum height (H), we use the formula H = Vy² / (2 * g). Substituting the known values, H = (173.2 m/s)² / (2 * 9.8 m/s²) ≈ 1532.87 m.