A ball being dropped and bouncing back up?

  • Thread starter Thread starter needlottahelp
  • Start date Start date
  • Tags Tags
    Ball
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a ball dropped from a height of 30 meters and its subsequent rebound to a height of 20 meters. Participants are exploring concepts related to kinematics and forces, particularly focusing on the average force exerted on the ball during its contact with the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the speed of the ball upon impact and the speed at which it rebounds. There are attempts to apply kinematic equations to find these speeds, and questions arise regarding the calculation of acceleration and the correct application of signs in the equations.

Discussion Status

Some participants have provided hints and guidance on the methods to approach the problem, including the use of kinematic equations and the relationship between speed, acceleration, and force. There is an acknowledgment of the need to be cautious with signs in calculations, but no explicit consensus or resolution has been reached.

Contextual Notes

Participants express uncertainty about the initial steps and the relevant equations, indicating a lack of familiarity with the material. The discussion is framed within the constraints of a homework assignment, which may limit the depth of exploration.

needlottahelp
Messages
29
Reaction score
0

Homework Statement


After falling from rest at a height of 30m, a 0.50 kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0 ms, what a average force was exerted on the ball?

Homework Equations


The kinematic equations?
F=ma
w=mg?
I don't really know

The Attempt at a Solution


Sorry I don't even know where to begin. I don't really understand this section in the book so i don't even know what to do. The answer is supposed to come out to 1.1 x 10^4 upward.
 
Last edited:
Physics news on Phys.org
You need to work out the speed the balls hits the floor and then the speed it starts backup at. Acceleration is just change in speed / time.

Hint - the speed it starts back up is the same as the speed a ball dropped from the second height would reach the floor.
 
Ok so the speed of the ball when it hits the ground first time would need the kinematic equation Vf^2= Vi^2 + 2ad right?

Vf^2= (0)^2 + 2(9.8)(30)
Vf^2 = 588
Vf~24.3 m/s ?

Vf^2= 2(9.8)(20)
Vf^2 = 392
Vf ~ 19.8 m/s ?

I think that's correct. correct me if I'm wrong please. How do I find the acceleration?
 
Acceleration = change in velocity / time
Whats the total change in velocity (watch the signs!)
 
Ohhh...

So the acceleration is the change in velocity which would be 19.8 - 24.3/change in time.
Then put it in the F=ma and solve for F correct? or is there more?
 
Correct method - but be careful of the signs.
 
OK thank you very much. I was stuck on this problem for a while.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
34
Views
3K
  • · Replies 13 ·
Replies
13
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
7
Views
3K
Replies
5
Views
6K
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
7
Views
3K