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A ball of mass m. initially at rest, is dropped from a height of 5 meters. If the coefficient or restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately________ m/s (take g = 9.8 m/s2)
Correct answer is '8.91'. Can you explain this answer?
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A ball of mass m. initially at rest, is dropped from a height of 5 met...
As we know that
The coefficient of restitution  = 
The velocity at the time of hitting with floor is = 
Velocity of approaching = 
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A ball of mass m. initially at rest, is dropped from a height of 5 met...
Problem: Find the speed of the ball just before it hits the floor for the second time if dropped from a height of 5 meters with a coefficient of restitution of 0.9.

Given:
- Mass of the ball (m) = m
- Height from which the ball is dropped (h) = 5 meters
- Coefficient of restitution (e) = 0.9
- Acceleration due to gravity (g) = 9.8 m/s^2

Solution:
When the ball is dropped from a height of 5 meters, it has potential energy which is converted to kinetic energy as it falls. When it hits the ground, some of the kinetic energy is lost due to the impact and rebound. The coefficient of restitution represents the fraction of kinetic energy retained after the impact and rebound.

Let's assume that the speed of the ball just before it hits the ground for the first time is v1. Using conservation of energy, we can find v1 as follows:

- Potential energy at height h = Kinetic energy just before hitting the ground
- mgh = (1/2)mv1^2
- v1 = √(2gh)

Substituting the given values, we get:

- v1 = √(2 x 9.8 x 5) = √(98) = 9.9 m/s (approx)

When the ball hits the ground, it rebounds with a speed of ev1. Let's assume that the speed of the ball just before it hits the ground for the second time is v2. Using conservation of energy again, we can find v2 as follows:

- Kinetic energy just before hitting the ground = Potential energy at height h
- (1/2)mv2^2 = mgh
- v2 = √(2gh)

Substituting the given values, we get:

- v2 = √(2 x 9.8 x 5) = √(98) = 9.9 m/s (approx)

Therefore, the speed of the ball just before it hits the floor the second time is approximately 0.9 x 9.9 = 8.91 m/s (approx).

Answer: The speed of the ball just before it hits the floor the second time is approximately 8.91 m/s (approx).
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A ball of mass m. initially at rest, is dropped from a height of 5 meters. If the coefficient or restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately________ m/s (take g = 9.8 m/s2)Correct answer is '8.91'. Can you explain this answer?
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A ball of mass m. initially at rest, is dropped from a height of 5 meters. If the coefficient or restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately________ m/s (take g = 9.8 m/s2)Correct answer is '8.91'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A ball of mass m. initially at rest, is dropped from a height of 5 meters. If the coefficient or restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately________ m/s (take g = 9.8 m/s2)Correct answer is '8.91'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass m. initially at rest, is dropped from a height of 5 meters. If the coefficient or restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately________ m/s (take g = 9.8 m/s2)Correct answer is '8.91'. Can you explain this answer?.
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