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A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)
Correct answer is '27.32'. Can you explain this answer?
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A stone piece of small size is suspended by a string of length L (20 ...
Given:
- Length of the string (L) = 20 m
- Angle at which the string becomes slack (θ)
To find:
- Value of initial velocity (v0) required for the stone to pass through the point of suspension
Approach:
1. Calculate the height (h) at which the stone is suspended from the point of suspension.
2. Determine the maximum height (H) reached by the stone during its parabolic motion.
3. Use the principle of conservation of mechanical energy to find the value of v0.
Calculation:
1. Calculate the height (h) at which the stone is suspended from the point of suspension.
- The length of the string forms the hypotenuse of a right-angled triangle with height (h) and base (L).
- Using trigonometry, we can write h = Lcosθ.
2. Determine the maximum height (H) reached by the stone during its parabolic motion.
- At the maximum height, the vertical component of velocity becomes zero.
- Using the equation of motion, v^2 = u^2 - 2gh, where u is the initial vertical velocity, g is the acceleration due to gravity, and h is the height.
- At maximum height, u = 0, so v^2 = -2gh.
- The maximum height can be calculated as H = (v0^2sin^2θ)/(2g).
3. Use the principle of conservation of mechanical energy to find the value of v0.
- At the point of suspension, the stone is at height h and has potential energy mgh.
- At the maximum height, the stone has potential energy mgh and zero kinetic energy.
- Using the principle of conservation of mechanical energy, mgh = 0 - mgh + 0.5mv0^2.
- Simplifying the equation, we get v0^2 = 2gh.
- Substituting the value of h, v0^2 = 2gLcosθ.
- Solving for v0, we get v0 = √(2gLcosθ).
Substituting the given values:
- Length of the string (L) = 20 m
- Angle at which the string becomes slack (θ)
- Acceleration due to gravity (g) = 9.8 m/s^2
Using v0 = √(2gLcosθ), we can calculate the value of v0.
- Length of the string (L) = 20 m
- Angle at which the string becomes slack (θ)
To find:
- Value of initial velocity (v0) required for the stone to pass through the point of suspension
Approach:
1. Calculate the height (h) at which the stone is suspended from the point of suspension.
2. Determine the maximum height (H) reached by the stone during its parabolic motion.
3. Use the principle of conservation of mechanical energy to find the value of v0.
Calculation:
1. Calculate the height (h) at which the stone is suspended from the point of suspension.
- The length of the string forms the hypotenuse of a right-angled triangle with height (h) and base (L).
- Using trigonometry, we can write h = Lcosθ.
2. Determine the maximum height (H) reached by the stone during its parabolic motion.
- At the maximum height, the vertical component of velocity becomes zero.
- Using the equation of motion, v^2 = u^2 - 2gh, where u is the initial vertical velocity, g is the acceleration due to gravity, and h is the height.
- At maximum height, u = 0, so v^2 = -2gh.
- The maximum height can be calculated as H = (v0^2sin^2θ)/(2g).
3. Use the principle of conservation of mechanical energy to find the value of v0.
- At the point of suspension, the stone is at height h and has potential energy mgh.
- At the maximum height, the stone has potential energy mgh and zero kinetic energy.
- Using the principle of conservation of mechanical energy, mgh = 0 - mgh + 0.5mv0^2.
- Simplifying the equation, we get v0^2 = 2gh.
- Substituting the value of h, v0^2 = 2gLcosθ.
- Solving for v0, we get v0 = √(2gLcosθ).
Substituting the given values:
- Length of the string (L) = 20 m
- Angle at which the string becomes slack (θ)
- Acceleration due to gravity (g) = 9.8 m/s^2
Using v0 = √(2gLcosθ), we can calculate the value of v0.
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Community Answer
A stone piece of small size is suspended by a string of length L (20 ...
Suppose the string becomes slack at point P.
v2 = gL cosθ ...(ii)
Using energy conservation,
v2 = v02 - 2gL(1 + cos θ) ...(iii)
From equations (i), (ii) and (iii):
v02 = gL(2 + 3 cos θ) ...(iv)
Using equations of motion,
L sin θ = v cos θ × t ...(v)
-L cos θ = v sin θ × t1/2gt2 ...(vi)
tanθ = √2 ...(vii)
So,
= g × 20 × [2 + 1.372}1/2
= (200 × 3.732)1/2
= 27.32 m/s
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A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer?
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A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer?.
A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone piece of small size is suspended by a string of length L (20 m). The piece is given a horizontal velocity v0. The string becomes slack at some angle θ and the piece describes a parabola. The value of v0, if the piece passes through the point of suspension, should be (in m/s) (Up to 2 decimal places)Correct answer is '27.32'. Can you explain this answer?.
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