Colision with a wall/platform and coefficient of restitution

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A ball launched against a solid wall rebounds and falls 2 feet away, while a subsequent throw onto a soft platform results in a 0.83-foot distance. The coefficient of restitution needs to be calculated based on the rebound speeds, which are 4.65 ft/s without the platform and 1.93 ft/s with it. The method involves dividing the rebound speeds to determine the coefficient, although the data may seem inconsistent. Clarification is provided that the platform is soft, contributing to the reduced rebound distance. The discussion emphasizes the importance of demonstrating the correct calculation method, even if the values appear illogical.
kohlerb
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1. A ball is launched against a solid wall with straight trajectory and a velocity of Vo. It hits the wall on a 3ft height and it falls on the ground at a 2ft distance (by the left of the wall). Then, an elastic platform is positioned on the exact height of the collision of the ball, and it is throwed once again, with the same velocity. But now, it falls on the ground at a distance of 0,83ft.
a) Coefficient of restitution = ?
b) Vo = ?

g = 32,17; No drag;
My attempt was to calculate the velocity right after the wall colision, and right after the platform colision, using kinematics equations for velocity (free fall and horizontal movement), and then divided both to find the coefficient, but that doesn't sound right to me. And I have no idea on how to calculate the Vo.
 
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Hi kohlerb. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

With the elastic platform, the ball rebounds far less distance? That doesn't sound right. (Maybe the data has been accidently swapped around?)

If this were an exam question, you should still "go through the motions" for solving it, even though the data values won't give a logical answer. At least demonstrate that you know the correct method for determining coefficient of restitution.

What would the rebound speeds be, for each throw?
 
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NascentOxygen said:
Hi kohlerb. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

With the elastic platform, the ball rebounds far less distance? That doesn't sound right. (Maybe the data has been accidently swapped around?)

If this were an exam question, you should still "go through the motions" for solving it, even though the data values won't give a logical answer. At least demonstrate that you know the correct method for determining coefficient of restitution.

What would the rebound speeds be, for each throw?

I'm sorry, I wrote the question poorly, the platform is a "soft" platform, it really makes the ball rebounds less. But I still don't know if the way I calculated the coefficient is the correct way. Can I divide the two rebound speeds to find it?

Rebound speeds:

Without the platform: 4,65ft/s
With the platform: 1,93ft/s
 
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With one of the rebound speeds being that of an elastic collision, you should be able to divide them. As you have not shown how you calculated them, I can't say whether your working looks right or not.
 
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