How High Will a Golf Ball Bounce After Falling Down a 3m Staircase?

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A golf ball dropped from a height of 3 meters will bounce back to its original height after striking each stair, assuming all collisions are elastic and neglecting air resistance. The ball's vertical velocity is zero at the start, and it gains speed as it falls, bouncing back to the same height after each impact. The discussion highlights that regardless of the number of stairs, the final bounce height remains 3 meters. The calculations suggest that the height of each step can vary, but this does not affect the maximum bounce height achieved. Ultimately, the golf ball will bounce back to 3 meters after reaching the bottom of the staircase.
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Homework Statement



A golf ball bounces down a flight of steel stairs, striking each stair on the way down. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is 3.00 m, determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.

Homework Equations



Momentum before collision = momentum after collision


The Attempt at a Solution


I tried setting up equations, but only got as far as mgh = 1/2mv^2 because of conservation of ME.
 
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Parzival said:

Homework Statement



A golf ball bounces down a flight of steel stairs, striking each stair on the way down. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is 3.00 m, determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.

Homework Equations



Momentum before collision = momentum after collision


The Attempt at a Solution


I tried setting up equations, but only got as far as mgh = 1/2mv^2 because of conservation of ME.

You need to approximate here:

Although the golf ball is creeping forward down the stair case, we have to assume the motion approximates vertical motion - you might imagine you are actually dropping the ball onto a platform that becomes progressively lower between each bounce - 1 step down, 2 steps down, 3 steps down etc.
Although the Earth will be accelerating up while the Golf ball accelerates down [and vice verca] the ratio of masses between the Earth and a golf ball means we can ignore the motion of the earth.


When two masses have an elastic, head on collision, the two bodies move away from the centre of mass after collision at the same speed as they approach the centre of mass before collision.
If we ignore the speed of the Earth {see above} this means that if the Golf ball approaches a step at 1.5 m/s, it leaves the step at 1.5 m/s.
Now, if the Golf Ball falls from the highest level, to the first step, it will gain a speed V, under the influence of gravity, as it falls. Since it bounces back at speed V, it will regain its original height while the effects of gravity stop it prior to the next plunge.
It now falls to the second step and bounces back to the original height. Then the 3rd, the 4th, the 5th etc.

How many steps are there.

If each step is 20 cm high [meaning 15 steps to cover the 3.00 m height, the bounces total

20 + 40 + 60 + 80 + ... + 280 + 300 = 2400cm

However if the steps were only 10 cm high, meaning 30 of them in all, the bounce total would be

10 + 20 + 30 + ... + 280 + 290 + 300 = 4650 cm

SO how many steps are there?

Perhaps the question is just after the final bounce height - which is 300 cm regardless of the number of stairs.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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