The marble will land approximately 2.35 meters away from the launch point.
How far away will the marble land if friction is neglected?
The initial velocity of 7 m/s needs to be broken down into its horizontal and vertical components.
Horizontal component (Vx): Vx = 7 m/s * cos(55°) ≈ 3.89 m/s
Vertical component (Vy): Vy = 7 m/s * sin(55°) ≈ 5.93 m/s
Since the floor is level, the horizontal component of the velocity (3.89 m/s) remains constant throughout the flight.
The vertical motion is governed by gravity (acceleration due to gravity = 9.81 m/s²).
Using the vertical component of the initial velocity and gravity, we can find the time the marble spends in the air (t):
Vy = V0y - gt
0 = 5.93 m/s - 9.81 m/s² * t
t ≈ 0.604 seconds
Range = Vx * t
Range ≈ 3.89 m/s * 0.604 seconds
Range ≈ 2.35 meters
Therefore, the marble will land approximately 2.35 meters away from the launch point.