Is Velocity Zero for Only an Instant at the Peak of a Ball's Trajectory?

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Discussion Overview

The discussion revolves around the behavior of a ball thrown upwards, specifically addressing the nature of its velocity at the peak of its trajectory and the calculation of acceleration based on given graphs of position and velocity. Participants explore both theoretical and practical aspects of motion under gravity, including the implications of constant acceleration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the velocity of a ball at the peak of its trajectory is zero for only an instant or for a longer duration, suggesting that the velocity graph is continuous and slopes downwards.
  • Another participant asserts that the velocity is zero only for an instant, particularly emphasizing that this applies to a ball thrown straight up, while noting that the vertical component is zero for an instant in a parabolic trajectory.
  • A participant expresses uncertainty about the definition of acceleration and whether it remains valid for a ball thrown upwards and downwards, questioning if acceleration could be constant throughout the motion.
  • One participant provides a hypothetical scenario where a ball thrown upwards decelerates due to gravity, concluding that it only stops for an instant after a specific time period.
  • Another participant clarifies that the velocity graph's behavior indicates the ball is rising when above the x-axis and falling when below, reinforcing that the acceleration remains constant despite the change in direction of velocity.

Areas of Agreement / Disagreement

Participants express differing views on the duration of zero velocity at the peak of the trajectory, with some asserting it is only for an instant while others suggest it may be longer. There is also uncertainty regarding the constancy of acceleration throughout the motion, indicating that multiple competing views remain unresolved.

Contextual Notes

Participants reference the definition of acceleration as the slope of the velocity graph, but there is ambiguity regarding its application in this context, particularly under varying conditions of motion. The discussion does not resolve the implications of these definitions on the calculations involved.

Who May Find This Useful

This discussion may be of interest to students and enthusiasts of physics, particularly those exploring concepts of motion, gravity, and the mathematical relationships between position, velocity, and acceleration.

physics213
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Two questions:

1.) When you throw a ball up and it reaches the height of its trajectory, is the velocity equal to zero only for an INSTANT or for LESS THAN A SECOND BUT MORE THAN AN INSTANT? The velocity graph is sloping straight downwards with no breaks.

2.) If you are given a x(t) graph that shows a simple parabolic motion (throw ball up and comes down) and a v(t) graph that is a straight line sloping downwards (starting above the x-axis and ending below the x axis), how would you calculate the acceleration (ball is thrown on another planet)?

Any help is appreciated. Thanks.
 
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Welcome to PF!

Hi physics213! Welcome to PF! :smile:
physics213 said:
1.) When you throw a ball up and it reaches the height of its trajectory, is the velocity equal to zero only for an INSTANT or for LESS THAN A SECOND BUT MORE THAN AN INSTANT? The velocity graph is sloping straight downwards with no breaks.

Only an instant (in other words, no time at all). :smile:

(Obviously, that's for a ball going straight up. If it's a parabola, then the vertical component is zero for an instant.)
2.) If you are given a x(t) graph that shows a simple parabolic motion (throw ball up and comes down) and a v(t) graph that is a straight line sloping downwards (starting above the x-axis and ending below the x axis), how would you calculate the acceleration (ball is thrown on another planet)?

Hint: what is the definition of acceleration? :smile:
 
tiny-tim said:
Hi physics213! Welcome to PF! :smile:


Only an instant (in other words, no time at all). :smile:

(Obviously, that's for a ball going straight up. If it's a parabola, then the vertical component is zero for an instant.)


Hint: what is the definition of acceleration? :smile:


I know the definition of acceleration is the slope of the v(t) graph, but does that definition still hold true for a ball going up and coming down? And that acceleration would be negative right? It just doesn't seem like the acceleration would be constant all the way through.
 
1)
For simplicity, imagine the object is thrown upwards at a rate of exactly g. After one second, it will "decelerate" by exactly acceleration due to gravity (1sec*g=g.) -- that is, after one second it will come to a complete stop. Now what happens if we wait 1.001 seconds? Obviously it won't be EXACTLY at a stand still. Hence it only stops for an instant.
 
physics213 said:
I know the definition of acceleration is the slope of the v(t) graph, but does that definition still hold true for a ball going up and coming down? And that acceleration would be negative right?
It just doesn't seem like the acceleration would be constant all the way through.

When the line of the velocity graph is above the x-axis, the ball is rising.

When it is below the x-axis, the ball is falling.

But it's an unbroken line … it goes through the x-axis "without blinking"!

The acceleration, as measured on the line, is the same whether the velocity is positive or negative … which matches the constant acceleration which you know gravity has. :smile:
 

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