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June 12th, 2012, 07:48 AM   #1
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A golf ball is dropped from a height of 10 m on a fixed stee

Hi all

A golf ball is dropped from a height of 10 m on a fixed steel plate. The coefficient of restitution is 0.894 . Find the height to which the ball rebounds on the fist ,second and third bounces ?
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June 12th, 2012, 08:58 AM   #2
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Re: A golf ball is dropped from a height of 10 m on a fixed

The kinetic energy of the ball when it hits the ground equals the potential energy it had when it was stationary at height H.

mv = mgH
or, dividing both sides by m:
v = gH

At each bounce, the new speed of the ball is 0.894 times the old speed.
Therefore the new kinetic energy is m(0.894 * v), which at the top of the bounce (height h) will have turned into a new potential energy of mgh.

(0.894 * v) = gh
But the left-hand side is 0.894 times v, which was equal to gH. Therefore:
0.894 * gH = gh
Dividing both sides by g:
h = 0.894 * H
h = 0.799236 * H
And this gives you the relationship between successive bounce heights.
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June 12th, 2012, 11:25 AM   #3
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Re: A golf ball is dropped from a height of 10 m on a fixed

this solving for teacher I don't understand clearly can please explain step by step

u1 =root(2gh)
velocity of plate before and after impact
u2= v2 = 0
e(u1 - u2 ) =V2-V1
e(root(2gh) - 0 ) = ( 0 - V1)
V1 = -e X root(2gh)
V^2 - u^2 = 2gh now u = zero
so
V1 = 2gh1
h1 = vb^2/2g
h1 = e^22gh/2g
h = e^2 X H (from this can find each height )
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June 13th, 2012, 05:04 AM   #4
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Re: A golf ball is dropped from a height of 10 m on a fixed

Quote:
Originally Posted by r-soy
u1 =root(2gh) [color=#00BF00]formula for velocity after falling from height h[/color]
velocity of plate before and after impact
u2= v2 = 0
e(u1 - u2 ) =V2-V1 [color=#00BF00]relationship between old/new velocities, with e as the coefficient of restitution[/color]
e(root(2gh) - 0 ) = ( 0 - V1)
V1 = -e X root(2gh)
V^2 - u^2 = 2gh now u = zero [color=#FF0000]Where did this come from?[/color]
so
V1 = 2gh1
h1 = vb^2/2g [color=#FF0000]I think vb means v1, and h1 is the new height[/color]
h1 = e^22gh/2g
h = e^2 X H (from this can find each height )
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June 14th, 2012, 05:43 AM   #5
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Re: A golf ball is dropped from a height of 10 m on a fixed

u1 =root(2gh) formula for velocity after falling from height h

why it not zero ,,, after falling it sould hit the second body

and I don't now from where he got V^2 - u^2 = 2gh
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