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velocity of separation = e (velocity of approach)

A little more explanation please?

|dw:1347425035783:dw|

does it make sense?

|dw:1347424692991:dw|

why are you subtracting 0?

Velocity of approach = V' - zero...which is velocity of ground

sorry...Velocity of approach = V - zero...which is velocity of ground

ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)

similarly velocity of separation = V'-0

okay. so e is the math constant..
what am i solving for?

so 1st calculate V, the velocity of ball from which it hits the ground 1st time..

no...e is the coefficient of restitution...and for ur problem e =0.85(given)

but see 4th line in ur solution..

The ball will bounce an infinite number of times, but the process will have a finite duration.

http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html
check this

So our professor gave us the wrong equation... awesome.

where are you from?

I'm studying in New Zealand.

ok

anyway...u can solve this problem by using correct formula

|dw:1347427306605:dw|

t' = time of reaching at the peak after 1st collision

does it make sense?

So far yes.

|dw:1347427668995:dw|

good...:)

e, e^2, e^3...will keep decreasing on...

general formula
\[v_n=\mu_r^n \sqrt{2gh}\]\[t_n=\mu_r^n \sqrt{2h \over g}\]
I need to find n.

yes...now u can tell the answer :)

the ball will stop when v(n)=0...right?

and restitution <1..so tell me the answer?

the ball will stop when v(n)=0

v(n)=0
\[0=\mu_r^n \sqrt{2gh}\] Right?

but now the question is when v(n) will be zero?

yes...are u noticing something?
e, e^2, e^3... with each v(n) for n=1,2,3,4....

and e=0.85 which is less than 1

now tell me... e^n will approach zero when n approaches....?

infinity

yes...perfectly right:)

so the ball will bounce infinite times...

well i got that answer like 3 hours ago... but i figured that there would be an actual answer.

ok

but practically ball can't go up to a size of atom after some bounces for large n

What you CAN work out is not number of bounces (infinite), but total time taken.

@Vincent-Lyon.Fr
Out of curiosity how would you go about doing that?