Calculating Acceleration and Velocity of a Falling Ball in One Dimension

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Homework Help Overview

The discussion revolves around calculating the acceleration and velocity of a ball that is dropped from a height of 3 meters and rebounds to a height of 2 meters. The problem involves concepts from kinematics, particularly focusing on the motion of the ball during its fall and subsequent bounce.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the speed of the ball at various points during its motion. Some participants question the appropriateness of using certain formulas over the bounce, suggesting a need for clarification on the application of constant versus non-constant acceleration equations.

Discussion Status

The discussion is active, with participants engaging in clarifying the application of formulas. There is an exchange of ideas regarding the correct approach to the second question about the speed of the ball after it leaves the ground, indicating a productive exploration of the topic.

Contextual Notes

Participants are navigating the complexities of kinematic equations in the context of a bouncing ball, with some uncertainty regarding the assumptions made in the calculations. There is a mention of an image attachment that includes formulas for non-constant acceleration, which may provide additional context but is not elaborated upon in the discussion.

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A ball is released from the height 3 meters and after it hits the floor it reaches the height 2 meter. A) Whats the speed of the ball in the moment when it meets the ground?
My answer : V^2-V0^2=2gs and here we find V=sqrt60.
What is the speed of the ball in the moment it leaves the ground? V2^2-V^2=2gh2 so V2^2- sqrt60=2g*2 and here we find V2=10
c) If the ball meets the ground for 0.02 seconds what's the direction and numerical value of the acceleration? The direction is upside down and a=(10-sqrt60)/0.02 ARE THESE CORRECT?
 
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Hi Elaia06! :smile:

(try using the X2 button just above the Reply box :wink:)
Elaia06 said:
A ball is released from the height 3 meters and after it hits the floor it reaches the height 2 meter. A) Whats the speed of the ball in the moment when it meets the ground?
My answer : V^2-V0^2=2gs and here we find V=sqrt60.

fine :smile:
What is the speed of the ball in the moment it leaves the ground? V2^2-V^2=2gh2 so V2^2- sqrt60=2g*2 and here we find V2=10

no, that's a constant acceleration formula, so you can't use it "over the bounce" …

try it with final speed 0 :wink:
 
You mean I should correct the third question,not the second one? maybe you did a little mistake over there?
 
no i did mean the second question (What is the speed of the ball in the moment it leaves the ground?)
 
I attached an image to the reply,it has the formulas for the non constant acceleration..but I don't have the time...
 

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