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A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)
Correct answer is '9'. Can you explain this answer?
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Verified Answer
A ball of mass m , initially at rest , is dropped from a height of 5 m...
9
Given that height = 5m
g = 10 m/sec2
Then the velocity of ball before hitting the surface first time =
coefficient of restitution = 0.9
So, after collision its velocity = 0.9 x 10 = 9 m/sec
So, before hits the floor second time velocity in = 9 m/sec
speed of object =
Given that height = 5m
g = 10 m/sec2
Then the velocity of ball before hitting the surface first time =
coefficient of restitution = 0.9
So, after collision its velocity = 0.9 x 10 = 9 m/sec
So, before hits the floor second time velocity in = 9 m/sec
speed of object =
Most Upvoted Answer
A ball of mass m , initially at rest , is dropped from a height of 5 m...
Given data:
Mass of the ball, m = ?
Height from which the ball is dropped, h = 5 meters
Coefficient of restitution, e = 0.9
Acceleration due to gravity, g = 10 m/s^2
Key Formula:
The velocity of an object just before hitting the ground can be calculated using the formula:
v = √(2gh)
where,
v is the velocity of the object just before hitting the ground,
g is the acceleration due to gravity,
and h is the height from which the object is dropped.
Explanation:
When the ball is dropped from a height of 5 meters, it falls due to the gravitational force acting on it. As it falls, it gains kinetic energy and its potential energy decreases. When the ball hits the ground, some of its kinetic energy is lost due to the collision with the ground.
The coefficient of restitution, e, represents the ratio of the final velocity to the initial velocity after a collision. In other words, it describes how elastic or how much energy is conserved during a collision.
In this case, the ball is dropped from rest, so the initial velocity is 0 m/s. The ball hits the ground and bounces back up, reaching a maximum height before falling back down again.
Using the formula mentioned above, we can calculate the velocity just before hitting the ground for the first time:
v1 = √(2gh)
= √(2 * 10 * 5)
= √(100)
= 10 m/s
Since the coefficient of restitution is 0.9, the final velocity after the first collision is:
v2 = e * v1
= 0.9 * 10
= 9 m/s
Now, the ball starts falling from the maximum height reached during the bounce. Using the same formula, we can calculate the velocity just before hitting the ground for the second time:
v3 = √(2gh)
= √(2 * 10 * 5)
= √(100)
= 10 m/s
Therefore, the speed of the ball just before it hits the floor for the second time is approximately 9 m/s.
Mass of the ball, m = ?
Height from which the ball is dropped, h = 5 meters
Coefficient of restitution, e = 0.9
Acceleration due to gravity, g = 10 m/s^2
Key Formula:
The velocity of an object just before hitting the ground can be calculated using the formula:
v = √(2gh)
where,
v is the velocity of the object just before hitting the ground,
g is the acceleration due to gravity,
and h is the height from which the object is dropped.
Explanation:
When the ball is dropped from a height of 5 meters, it falls due to the gravitational force acting on it. As it falls, it gains kinetic energy and its potential energy decreases. When the ball hits the ground, some of its kinetic energy is lost due to the collision with the ground.
The coefficient of restitution, e, represents the ratio of the final velocity to the initial velocity after a collision. In other words, it describes how elastic or how much energy is conserved during a collision.
In this case, the ball is dropped from rest, so the initial velocity is 0 m/s. The ball hits the ground and bounces back up, reaching a maximum height before falling back down again.
Using the formula mentioned above, we can calculate the velocity just before hitting the ground for the first time:
v1 = √(2gh)
= √(2 * 10 * 5)
= √(100)
= 10 m/s
Since the coefficient of restitution is 0.9, the final velocity after the first collision is:
v2 = e * v1
= 0.9 * 10
= 9 m/s
Now, the ball starts falling from the maximum height reached during the bounce. Using the same formula, we can calculate the velocity just before hitting the ground for the second time:
v3 = √(2gh)
= √(2 * 10 * 5)
= √(100)
= 10 m/s
Therefore, the speed of the ball just before it hits the floor for the second time is approximately 9 m/s.
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A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. Can you explain this answer?
Question Description
A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. 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 of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. 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 of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. Can you explain this answer?.
A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. 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 of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. 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 of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. 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 of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. Can you explain this answer?, a detailed solution for A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. Can you explain this answer? has been provided alongside types of A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)Correct answer is '9'. Can you explain this answer? theory, EduRev gives you an
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