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BodhiAI
Nov. 2, 2019

Bouncing Of Ball:

Rebounding of Ball After Collision With Ground :-

If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.

 

 

v1 = ev0 = e√2gh0

As e= velocity after collision/velocity before collision

(1) First height of rebound :

h1 = v21/2g = e2h0

  h1 = e2h0

(2) Height of the ball after nth rebound: Obviously, the velocity of ball after nth rebound will be

vn = e2h0

Therefore the hight after nth rebound will be

Hn = v2n/2g = e2h0

  hn = e2nh0

Total distance travelled by the ball before it stops bouncing

H = h0 + 2h1 + 2h2 + 2h3 + ….= h0 + 2e1h0 + 2e4h0+ 2e6h0+ …

H = h0 [1 + 2e2(1+ e2 + e4+ e6…….)]

= h0 1+ 2e2 /(1-e2)

As 1+ e2+e4+ ….=1/(1-e2)

  H = h0 = (1+e2)/(1-e2)

Total time taken by the ball to stop bouncing

T = h0 + 2t1 + 2t2 + 2t3 + ….= (2h0/g) + 2(2h1/g) + 2(2h2/g) +…..

=2h0/g [1+2e+2e2 +……….] [As h1 = e2h0; h2 = e4h0]

=√2h0/g [1+2e(1 + e + e2 + e3 +….)]

=√(2h0/g)[ 1+2e/(1-e)] = (2h0/g )[(1+e)/(1-e)]

T = (1+e/1-e) √(2h0/g)