What is the Correct Calculation for h/H When Ball A Collides with Ball B?

AI Thread Summary
In the discussion about the collision between Ball A and Ball B, it is established that Ball A is dropped from height H while Ball B is thrown upwards. At the moment of collision, Ball A has twice the speed of Ball B, leading to confusion regarding the correct relationship between their speeds. The participant realizes that the speed of Ball A should be represented as 2 instead of 4 in their calculations. They propose an alternate method using relative velocity and simultaneous equations to solve for height h in relation to H. The discussion emphasizes the importance of accurately interpreting speed relationships in collision problems.
Juliusz
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Homework Statement


Ball A is dropped from rest from a building of height H exactly as ball B is thrown up vertically from the ground. When they collide A has twice the speed of B. If the collision occurs at height h, what is h/H?

Homework Equations


L9rtuun.png

(I would type them out put I can't format them properly, sorry)

The Attempt at a Solution


ZH9mqyG.png


I did everything as in the example, but then I got to the part where we compare the speeds. The formula (red arrow) show that at the time of impact, the speed of A is 4 times greater than speed of B. But in the problem, it states that the speed of A is only twice the speed of B. So my understanding is that there should be a 2 instead of a 4. Am I missing something?
 

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Juliusz said:
Am I missing something?

The equation is in ##v^2##.
 
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Ah, I think I see. So it can be written like this?
GJDOdpq.png
 

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Alternate method based on relative velocity of A and B as well as the speed relation when they collide:

Solve simultaneously for H and t (in terms of u - initial velocity of B - and g):

t = H/u ......Eqn 1
2(u-gt) = gt ....Eqn 2

Then h = ut - 0.5 g t^2.
 
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