Final answer:
The maximum height reached by the cannonball is approximately 31.9 meters, and it lands approximately 220.3 meters away from the launch point using the principles of projectile motion.
Step-by-step explanation:
To determine the maximum height and the horizontal distance the cannonball lands, we use the principles of projectile motion.
Maximum Height
The vertical component of the initial velocity (Vy) can be calculated using Vy = V * sin(θ), where V is the initial velocity and θ is the launch angle. The formula Vy = 50 m/s * sin(30°) gives us Vy = 25 m/s. We then use the equation h = Vy^2 / (2 * g), where g is the acceleration due to gravity (9.81 m/s^2), to find the maximum height h. Plugging in the values, we find h = 25^2 / (2 * 9.81) ≈ 31.9 m.
Horizontal Distance
For horizontal distance (range), we use the horizontal component of the initial velocity (Vx), which is Vx = V * cos(θ). So Vx = 50 m/s * cos(30°) ≈ 43.3 m/s. The time of flight (t) is given by t = (2 * Vy) / g. Using our Vy, we get t ≈ 5.1 s. The range R can be found using R = Vx * t, which gives us R ≈ 220.3 m.