+0  
 
0
1050
4
avatar+426 

 

 Apr 27, 2016

Best Answer 

 #3
avatar+37146 
+5

Before the ball is dropped it has POTENTIAL energy = mgh    all of this energy is converted to kintetic energy 1/2 mv^2 at the point where the basketball hits the floor

(original) mgh = 1/2 mv^2(at floor)        I cannot graph these equations here.    g'luck !

 Apr 27, 2016
 #1
avatar+118673 
+5

         

This is MWizzard's physics question.  It is not answered yet

 

 

 

 

 Apr 27, 2016
 #2
avatar+37146 
+5

For 'B'   it is not a perfectly elastic collision between the ball and the concrete....   some of the energy is lost in making a sound (air vibrations) some of the enrgy is lost in vibrations in the concrete, some of the nergy is lost in deformation of the ball AND the concrete...even though they may spring back to the original shape, this deformation results in heat in the ball and the concrete,  lastly , thee is AIR friction to which the ball loses energy...this results in a minute heating of th air and the ball.

 Apr 27, 2016
 #3
avatar+37146 
+5
Best Answer

Before the ball is dropped it has POTENTIAL energy = mgh    all of this energy is converted to kintetic energy 1/2 mv^2 at the point where the basketball hits the floor

(original) mgh = 1/2 mv^2(at floor)        I cannot graph these equations here.    g'luck !

ElectricPavlov Apr 27, 2016
 #4
avatar+33661 
+5

Here are a couple of relevant graphs:

 

bouncing

.

 Apr 27, 2016

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