First, we have to lay down the given data: the angle of dive is 55.0 degrees, the altitude at the time of release is 570 m, and the time taken by the projectile to hit the ground is 8.00 s.
Let's start off by figuring out the vertical distance that the projectile has travelled in this time. We use the equations of motion here, which in the case of free fall is described by the equation d = 0.5 * g * t^2, where d is the distance travelled, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken. Substituting the given values into this equation, we can find the total vertical distance that the projectile has travelled.
However, we need to remember that the projectile was not released from ground level, but from an altitude of 570 m. This means that the actual vertical distance travelled by the plane is the total vertical distance that the projectile travelled, minus the given altitude.
Now, let's consider the right triangle formed by the vertical and horizontal movement of the plane. The resultant path of the plane (the hypotenuse of this triangle) is given by the equation plane's actual distance = plane's vertical distance / sin(angle). Before we can use this equation, we will need to convert the angle from degrees to radians.
Once we have found the plane's actual distance travelled, we can use the basic definition of speed to find the speed of the plane. Speed is simply distance over time, so by dividing the plane's actual distance by the time taken (8.00 s), we can find the speed of the plane.
By following these steps, we find that the speed of the aircraft is approximately -39.13 m/s. The negative sign merely indicates the direction of the aircraft's movement, which is downward and to the right in our chosen coordinate system.