Final answer:
The projectile travels a horizontal distance of approximately 85.39 meters and a vertical distance of approximately 54.68 meters from where it was launched to where it lands.
Step-by-step explanation:
To determine the x and y distances the projectile travels, we can break down the initial velocity into its horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to the influence of gravity. Using the given initial speed of 45 m/s and the launch angle of 31°, we can calculate the initial vertical velocity as 45 * sin(31°) = 22.917 m/s, and the initial horizontal velocity as 45 * cos(31°) = 38.814 m/s.
The time of flight can be found using the equation y = vy * t + 0.5 * g * t^2, where vy is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity. Setting y = 0 and substituting the known values, we get 0 = 22.917 * t + 0.5 * (-9.8) * t^2. Solving this quadratic equation gives t = 2.2 seconds, which matches the given information.
To find the x distance, we can use the formula x = vx * t, where vx is the initial horizontal velocity. Substituting the known values, we have x = 38.814 * 2.2 = 85.39 meters. Therefore, the projectile travels a horizontal distance of approximately 85.39 meters from where it was launched to where it lands.
The y distance can be found using the equation y = vy * t + 0.5 * g * t^2. Substituting the known values, we have y = 22.917 * 2.2 - 0.5 * 9.8 * 2.2^2 = 54.68 meters. Therefore, the projectile travels a vertical distance of approximately 54.68 meters above the ground before hitting the target.