Final answer:
The correct answer is B) The horizontal velocity Vx = 15 m/s while the vertical velocity Vy = 0 m/s. This is because, at the highest point of a projectile's trajectory, the vertical velocity is 0 m/s and the horizontal velocity remains constant as there is no horizontal acceleration.
Step-by-step explanation:
The student is asking about the conditions at the highest point of the trajectory of a projectile launched at a 45-degree angle. At this point, a couple of things can be noted about the projectile's velocity and acceleration:
- The vertical velocity (Vy) is 0 m/s because the projectile is momentarily stationary in the vertical direction before starting to fall back down.
- The vertical acceleration (Ay) is equal to the acceleration due to gravity, which is -9.8 m/s². This is consistent in projectile motion regardless of its position along its trajectory.
- The horizontal velocity (Vx) remains constant throughout the flight because there is no acceleration in the horizontal direction in ideal projectile motion (assuming no air resistance).
- Both horizontal and vertical acceleration cannot be zero because gravity is always acting on the projectile, giving it a vertical acceleration of -9.8 m/s².
Therefore, the correct answer to the student's question is B) The horizontal velocity Vx = 15 m/s while the vertical velocity Vy = 0 m/s.
Including the example from the information provided, at the peak of its trajectory, a football punted at a 45.0° angle with no wind effect, would have a horizontal velocity that remains unchanged from launch (assuming no air resistance), and the vertical velocity would be 0 m/s. The vertical acceleration due to gravity would remain constant at -9.8 m/s². The horizontal acceleration remains 0 m/s², because, in ideal projectile motion, there are no forces acting in the horizontal direction to change the velocity.