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A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :
- a)30º
- b)45º
- c)60º
- d)120º
Correct answer is option 'A'. Can you explain this answer?
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A body is projected with a speed u m/s at an angle β with the hor...
K.E. = K cos2 β
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A body is projected with a speed u m/s at an angle β with the hor...
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A body is projected with a speed u m/s at an angle β with the hor...
Given data:
- Initial speed of projection (u)
- Angle with the horizontal (β)
- Kinetic energy at the highest point is 3/4th of the initial kinetic energy
Understanding the problem:
- At the highest point, the body has maximum potential energy and zero kinetic energy.
- So, kinetic energy at the highest point is equal to the initial kinetic energy minus the potential energy at the highest point.
Solution:
- Let the initial kinetic energy be K.
- At the highest point, kinetic energy = 3/4 * K
- Potential energy = K - 3/4 * K = 1/4 * K
- At the highest point, kinetic energy = potential energy
- Using trigonometry, we can relate the kinetic energy at the highest point to the initial kinetic energy.
Calculation:
- Let's assume the initial kinetic energy be K.
- Using the formula for kinetic energy (K = 1/2 * m * u^2), we can write the kinetic energy at the highest point as 1/2 * m * (u * sin(β))^2.
- Given that this kinetic energy is 3/4 of the initial kinetic energy, we have: 1/2 * m * (u * sin(β))^2 = 3/4 * K
- Simplifying the equation, we get: sin(β)^2 = 3/2
- Solving for β, we find that β = 30°.
Therefore, the value of β is 30° (Option A).
- Initial speed of projection (u)
- Angle with the horizontal (β)
- Kinetic energy at the highest point is 3/4th of the initial kinetic energy
Understanding the problem:
- At the highest point, the body has maximum potential energy and zero kinetic energy.
- So, kinetic energy at the highest point is equal to the initial kinetic energy minus the potential energy at the highest point.
Solution:
- Let the initial kinetic energy be K.
- At the highest point, kinetic energy = 3/4 * K
- Potential energy = K - 3/4 * K = 1/4 * K
- At the highest point, kinetic energy = potential energy
- Using trigonometry, we can relate the kinetic energy at the highest point to the initial kinetic energy.
Calculation:
- Let's assume the initial kinetic energy be K.
- Using the formula for kinetic energy (K = 1/2 * m * u^2), we can write the kinetic energy at the highest point as 1/2 * m * (u * sin(β))^2.
- Given that this kinetic energy is 3/4 of the initial kinetic energy, we have: 1/2 * m * (u * sin(β))^2 = 3/4 * K
- Simplifying the equation, we get: sin(β)^2 = 3/2
- Solving for β, we find that β = 30°.
Therefore, the value of β is 30° (Option A).
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A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer?.
A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer?.
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A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :a)30ºb)45ºc)60ºd)120ºCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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