Aprojectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0â° above the horizontal. 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 it lands?

x = 129.9 m y = 30.9 m First, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m