Question
A ball is thrown straight up into the air. Which of the following is true about the ball's velocity \( v \) and acceleration \( a \) at the highest point of its path? A. Both \( v=0 \) and \( a=0 \) B. \( v \neq 0 \) but \( a=0 \) C. \( v=0 \) but \( a \neq 0 \)
Ask by Hanson Thornton. in the United States
Feb 03,2025
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Tutor-Verified Answer
Answer
At the highest point, the ball's velocity is 0, but its acceleration is not 0.
Solution
At the highest point in the ball's trajectory, its behavior in terms of velocity and acceleration can be understood as follows:
1. **Velocity (\( v \))**:
- **Instantaneous Velocity**: At the very top of its path, the ball momentarily stops before changing direction from upward to downward. This means the instantaneous velocity \( v = 0 \) at that specific point.
2. **Acceleration (\( a \))**:
- **Constant Acceleration**: Regardless of the ball's position in its trajectory, the acceleration due to gravity (\( g \)) acts downward. This acceleration remains constant at approximately \( 9.8 \, \text{m/s}^2 \) (neglecting air resistance) throughout the ball's flight, including at the highest point.
Combining these insights:
- **Velocity at Highest Point**: \( v = 0 \)
- **Acceleration at Highest Point**: \( a \neq 0 \) (specifically, \( a = -g \) if we take downward as negative)
**Therefore, the correct statement is:**
**C. \( v = 0 \) but \( a \neq 0 \)**
Reviewed and approved by the UpStudy tutoring team
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Beyond the Answer
At the highest point of its path, the ball momentarily comes to a stop, meaning that its velocity \( v \) is indeed 0. However, gravity is still acting on the ball, which means its acceleration \( a \) is not 0; it is equal to -9.81 m/s² (downward). So, the correct answer is C: \( v=0 \) but \( a \neq 0 \). If you're ever in doubt about your physics, just remember that while objects may momentarily stop at their peak, forces like gravity are always in action, ready to bring them back down! Stay curious about how motion works—there's always more to discover!