Final answer:
The horizontal displacement of the cannonball, shot at an angle of 60° with a velocity of 125 m/s, is calculated using the principles of projectile motion, resulting in an approximate displacement close to the given option (b) 1381.3 m.
Step-by-step explanation:
The problem involves calculating the horizontal displacement of a cannonball shot at an angle of 60° with an initial velocity of 125 m/s, a classical example of projectile motion. To solve this problem, we will use the kinematic equations of projectile motion. However, we need to decompose the initial velocity into horizontal and vertical components first. The horizontal component (Vx) is given by Vx = V * cos(θ), where V is the initial velocity and θ is the angle. Similarly, the vertical component (Vy) is given by Vy = V * sin(θ).
In this case, Vx = 125 * cos(60°) = 62.5 m/s and the time taken for the projectile to hit the ground is determined by the vertical component and the acceleration due to gravity, neglecting air resistance. The vertical motion forms a parabola, and the time it takes to reach the ground is twice the time it takes to reach the maximum height. We find the time to reach maximum height (t) using Vy = g*t, where g is the acceleration due to gravity (9.81 m/s²). Thus, t = Vy/g = (125 * sin(60°))/9.81. The total time in the air (δt), which is twice of t, gives us the total time of flight. Finally, the horizontal displacement (δx) is given by δx = Vx * δt.
Carrying out these calculations, we find that t = (125 * sin(60°)) / 9.81 ≈ 10.9 s, and therefore δt = 2 * t ≈ 21.8 s. Subsequently, δx = 62.5 m/s * 21.8 s ≈ 1363.8 m, which means the closest option is (b) 1381.3 m. There might be slight discrepancies due to rounding of the intermediate and final answers.