How Do I Calculate the Rebound Heights of a Dropped Golf Ball?

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To calculate the rebound heights of a golf ball dropped from 10 meters with a coefficient of restitution of 0.894, the initial velocity before impact is derived using the formula u1 = √(2gh), where g is the acceleration due to gravity. The coefficient of restitution relates the velocities before and after the impact, leading to the equation V1 = -e * u1, where V1 is the velocity after the bounce. The height of the first rebound can be calculated using h1 = V1² / (2g), which simplifies to h1 = e² * H, where H is the initial height. This process can be repeated for subsequent bounces to find the heights of the second and third rebounds. Understanding these equations and their derivations is essential for calculating the rebound heights accurately.
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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 ?

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 )

Now I want from you explain the solving In fact I don't understand the solving of my teacher
 
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Which part that you do not understand?
 
acuuly all parts I don't understand

why u = =root(2gh) why not zero

from where this equation V^2 - u^2 = 2gh

please can explain to me step by step
 
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