Final answer:
To calculate the time a bullet fired at a 45-degree angle with an initial velocity of 200 m/s is in the air, use the initial vertical velocity component and the formula t = (2u sin(θ))/g, yielding approximately 28.85 seconds.
Step-by-step explanation:
Finding the Time a Projectile Is in the Air
When a bullet is fired at an angle of 45 degrees with an initial velocity of 200 m/s, we can calculate the time the projectile is in the air by considering the vertical component of the velocity. Since the bullet is projected at a 45-degree angle, we can use trigonometry to find the initial vertical velocity to be 200 m/s × sin(45) = 141.42 m/s. The time spent in the air can be calculated using the formula for the time of ascent or descent under gravity, which is t = (2u sin(θ))/g, where u is the initial velocity, θ is the projection angle, and g is the acceleration due to gravity (9.8 m/s2).
Plugging in the values, we calculate t = 2 × 141.42 m/s / 9.8 m/s2 to obtain the time the bullet is in the air, which is approximately 28.85 seconds. This calculation assumes there is no air resistance and that the height from which the bullet is fired and lands is the same.