The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.. . Part C. . What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.. . A) 0, 0, 0, 0. B) 0, 0, 0, -9.80. C) 15.0, 0, 0, 0. D) 15.0, 0, 0, -9.80. E) 0, 26.0, 0, 0. F) 0, 26.0, 0, -9.80. G) 15.0, 26.0, 0, 0. H) 15.0, 26.0, 0, -9.80
Answers
Final answer:
At the peak of its trajectory, the ball's vertical velocity (v1,y) is zero, whereas its horizontal velocity (v1,x) maintains its consistency. The accelerations a1,x and a1,y are 0 and -9.8 m/s² respectively.
Explanation:
The physical concept analyzed here is projectile motion. Considering the trajectory of a ball, we can separate its motion into horizontal and vertical components. At the peak of its trajectory (time t1), the ball's vertical velocity (v1,y) is zero, because it temporarily stops moving upwards before it begins to fall again. However, since there is no acceleration in the horizontal direction (a1,x), the ball keeps moving horizontally with constant velocity (v1,x).
The acceleration in the vertical direction (a1,y) is still the acceleration due to gravity (-9.8 m/s²). This is because even at the peak of the trajectory, the ball is still accelerating downwards. Therefore, the correct answer should include v1,y = 0 m/s, v1,x = consistent value, a1,x = 0 m/s², and a1,y = -9.8 m/s². So the most probable answer is D) 15.0, 0, 0, -9.80, assuming the ball's horizontal velocity is 15 m/s. Please check the exact value of horizontal velocity (v1,x) in the settings of your problem.
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