What is the total displacement of a freely falling body, after successive rebounds from the same place of ground, before it comes to stop? Assume that 'e' is the coefficient of resolution between the body and the ground.
Answers
Answered by
9
Hey mate,
I'll explain this with simplest method.
◆ Answer - hn = h.(e^2)^n
● Explaination-
Suppose an object at height h rebounding from floor n number of times.
The coefficient of restitution is given by
e = √(h1/h)
h1 = e^2.h
Similarly,
h2 = e^2.h1 = (e^2)^2.h
h3 = e^2.h2 = (e^2)^3.h
...............
This is geometric progression with first term h and common ratio e^2.
From this, nth term is given by
hn = h.(e^2)^n
Hope that is helpful...
I'll explain this with simplest method.
◆ Answer - hn = h.(e^2)^n
● Explaination-
Suppose an object at height h rebounding from floor n number of times.
The coefficient of restitution is given by
e = √(h1/h)
h1 = e^2.h
Similarly,
h2 = e^2.h1 = (e^2)^2.h
h3 = e^2.h2 = (e^2)^3.h
...............
This is geometric progression with first term h and common ratio e^2.
From this, nth term is given by
hn = h.(e^2)^n
Hope that is helpful...
Answered by
2
Hi friends!
Assume that 'e' is the coefficient of restitution between the body and the ground.
If a body dropped freely from a certain height h, the displacement of the body is h.
The total distance covered=h(1+e^2)/ (1-e^2) , where h is height through which it has fallen and e is the coefficient of restitution.
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