Question

A projectile is launched from the groud with an angle of 30 degrees and an initial velocity of 50 m/s. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where is lands?

Answer 1

The x distance from where the projectile was launched to where it lands is 129.9 meters and the y distance is 37.5 meters.

Step-by-step explanation:

To find the x and y distances from where the projectile was launched to where it lands, we can break down the initial velocity into its x and y components. The x component (horizontal) can be found using the equation: Vx = initial velocity * cos(angle). Substituting the given values, we have Vx = 50 m/s * cos(30°) = 43.3 m/s. The y component (vertical) can be found using the equation: Vy = initial velocity * sin(angle). Substituting the given values, we have Vy = 50 m/s * sin(30°) = 25 m/s.

Since the projectile lands after 3 seconds, we can use the equation: vertical distance = Vy * time + 1/2 * acceleration due to gravity * time^2. Substituting the given values, we have vertical distance = 25 m/s * 3 s + 1/2 * 9.8 m/s^2 * (3 s)^2 = 37.5 m. Therefore, the x distance is equal to the horizontal distance traveled, which is horizontal distance = Vx * time = 43.3 m/s * 3 s = 129.9 m. So, the x distance from where the projectile was launched to where it lands is 129.9 meters and the y distance is 37.5 meters.