Final answer:
The projectile spends approximately 7.26 seconds in the air before hitting the ground, as determined by its initial vertical velocity, the effects of gravity, and the initial height from which it was fired.
Step-by-step explanation:
The total time a projectile spends in the air is determined by its vertical motion. For a projectile fired from a height with an initial velocity at an angle, we can decompose the initial velocity into horizontal and vertical components. The horizontal component affects how far the projectile will travel, while the vertical component, along with the effects of gravity, determines how long it will stay in the air.
In this case, the projectile is fired with an initial speed of 30 m/s at a 45° angle. The initial vertical velocity (vy) can be found using the sine of the angle: vy = v * sin(45°) = 30 m/s * 0.707 = 21.2 m/s. From this, we can calculate the time to reach the peak of the trajectory, and then double it to account for the projectile's ascent and descent before striking the ground. However, we also need to consider the additional time taken to fall the extra 200 m from the height of the cliff. The total time in the air (t) is given by the quadratic formula solving for the roots of the equation of motion in the vertical direction.
After calculations using the appropriate kinematic equations considering initial vertical velocity, acceleration due to gravity (-9.81 m/s2), and the initial height of the projectile, we find that the projectile spends approximately 7.26 seconds in the air before hitting the ground.