Final answer:
To find the range of the projectile launched at a 30-degree angle with an initial velocity of 10 m/s, calculate the horizontal and vertical velocity components, the time of flight, and then multiply the horizontal velocity component by the time of flight to obtain the range, which is approximately 8.83 meters.
Step-by-step explanation:
To calculate the range of this projectile motion, we use the following equations: The horizontal velocity component (Vx) is found using Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle. The vertical velocity component (Vy) is found using Vy = V * sin(θ). The time of flight (T) is found by calculating the time it takes for the projectile to rise and then return to the ground, using the formula T = (2 * Vy) / g, where g is the acceleration due to gravity (9.81 m/s^2). The range (R) can then be found using R = Vx * T. Using the values provided: Vx = 10 m/s * cos(30°) = 10 m/s * (√3/2) ≈ 8.66 m/s, Vy = 10 m/s * sin(30°) = 10 m/s * (1/2) = 5 m/s, T = (2 * 5 m/s) / 9.81 m/s^2 ≈ 1.02 s, R = 8.66 m/s * 1.02 s ≈ 8.83 m. Therefore, the range of the projectile is approximately 8.83 meters.