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 June 12th, 2012, 07:48 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 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 ? June 12th, 2012, 08:58 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 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. June 12th, 2012, 11:25 AM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 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 ) 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

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 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 ) June 14th, 2012, 05:43 AM #5 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 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 Tags ball, dropped, fixed, golf, height, stee ,
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# what is the COR of a ball dropped from 1m?

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