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A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone is

a. 8 m s−1
b. 15 m s−1
c. 12 m s−1
d. 9 m s−1?

a. 8 m s−1
b. 15 m s−1
c. 12 m s−1
d. 9 m s−1?
Most Upvoted Answer
A stone is projected from a point on the ground in such a way so as to...
Analysis:
- Let the initial speed of the stone be \( u \) and the angle of projection be \( \theta \).
- The horizontal component of the velocity of the stone remains constant at \( u\cos\theta \), as no horizontal force is acting on it.
- The vertical component of the velocity of the stone is given by \( u\sin\theta - gt \), where \( g \) is the acceleration due to gravity and \( t \) is the time taken for the stone to reach the bird at height \( 3h \).
- The time taken for the stone to reach the bird at height \( 3h \) can be calculated using the vertical motion equation: \( 3h = ut\sin\theta - \frac{1}{2}gt^2 \).
Solution:
1. Calculating time of flight:
- Using the above equation, we can find the time taken for the stone to reach the bird at height \( 3h \).
- \( 3h = u\sin\theta t - \frac{1}{2}gt^2 \).
2. Calculating horizontal distance:
- The horizontal distance covered by the bird in time \( t \) is \( 10t \) m.
- Since the stone hits the bird while descending, the horizontal distance covered by the stone is also \( 10t \) m.
- Therefore, \( 10t = u\cos\theta t \), which gives \( u = 10\cos\theta \) m/s.
3. Calculating vertical distance:
- At the maximum height of \( 4h \), the vertical component of the velocity of the stone is \( 0 \).
- Using this information, we can find the initial velocity of the stone, \( u \).
4. Calculating horizontal velocity:
- Now that we have the initial velocity of the stone, we can find the horizontal velocity using \( u = 10\cos\theta \) m/s.
Therefore, the horizontal velocity of the stone is 15 m/s (option b).
- Let the initial speed of the stone be \( u \) and the angle of projection be \( \theta \).
- The horizontal component of the velocity of the stone remains constant at \( u\cos\theta \), as no horizontal force is acting on it.
- The vertical component of the velocity of the stone is given by \( u\sin\theta - gt \), where \( g \) is the acceleration due to gravity and \( t \) is the time taken for the stone to reach the bird at height \( 3h \).
- The time taken for the stone to reach the bird at height \( 3h \) can be calculated using the vertical motion equation: \( 3h = ut\sin\theta - \frac{1}{2}gt^2 \).
Solution:
1. Calculating time of flight:
- Using the above equation, we can find the time taken for the stone to reach the bird at height \( 3h \).
- \( 3h = u\sin\theta t - \frac{1}{2}gt^2 \).
2. Calculating horizontal distance:
- The horizontal distance covered by the bird in time \( t \) is \( 10t \) m.
- Since the stone hits the bird while descending, the horizontal distance covered by the stone is also \( 10t \) m.
- Therefore, \( 10t = u\cos\theta t \), which gives \( u = 10\cos\theta \) m/s.
3. Calculating vertical distance:
- At the maximum height of \( 4h \), the vertical component of the velocity of the stone is \( 0 \).
- Using this information, we can find the initial velocity of the stone, \( u \).
4. Calculating horizontal velocity:
- Now that we have the initial velocity of the stone, we can find the horizontal velocity using \( u = 10\cos\theta \) m/s.
Therefore, the horizontal velocity of the stone is 15 m/s (option b).
Community Answer
A stone is projected from a point on the ground in such a way so as to...
Given Data:
- Height of the bird, h = 3h
- Maximum height attained by the stone, H = 4h
- Horizontal velocity of the bird, V = 10 m/s
Analysis:
- The stone is projected from the ground and attains a maximum height of 4h before hitting the bird at a height of 3h.
- The stone hits the bird while descending, which means the stone and bird are at the same height at the time of collision.
- The horizontal velocity of the stone remains constant throughout its motion.
Solution:
- Let the time taken by the stone to reach the maximum height be t1 and the total time taken to hit the bird be t.
- Using the equation of motion, H = ut + 1/2 * a * t^2, where u is the initial velocity and a is the acceleration due to gravity.
- At the maximum height, the vertical component of the velocity becomes zero, so we have 0 = u * sin(theta) - g * t1, where theta is the angle of projection.
- Solving the above equations, we can find the values of t1 and t.
Calculation:
- Let t1 be the time taken to reach the maximum height and t be the total time taken to hit the bird.
- Using the equation H = ut + 1/2 * a * t^2, we get 4h = u * cos(theta) * t1 - 1/2 * g * t1^2 and 3h = u * cos(theta) * t - 1/2 * g * t^2.
- By solving the above equations, we can find the value of t.
Final Answer:
- The horizontal velocity of the stone is 12 m/s. (Option c)
- Height of the bird, h = 3h
- Maximum height attained by the stone, H = 4h
- Horizontal velocity of the bird, V = 10 m/s
Analysis:
- The stone is projected from the ground and attains a maximum height of 4h before hitting the bird at a height of 3h.
- The stone hits the bird while descending, which means the stone and bird are at the same height at the time of collision.
- The horizontal velocity of the stone remains constant throughout its motion.
Solution:
- Let the time taken by the stone to reach the maximum height be t1 and the total time taken to hit the bird be t.
- Using the equation of motion, H = ut + 1/2 * a * t^2, where u is the initial velocity and a is the acceleration due to gravity.
- At the maximum height, the vertical component of the velocity becomes zero, so we have 0 = u * sin(theta) - g * t1, where theta is the angle of projection.
- Solving the above equations, we can find the values of t1 and t.
Calculation:
- Let t1 be the time taken to reach the maximum height and t be the total time taken to hit the bird.
- Using the equation H = ut + 1/2 * a * t^2, we get 4h = u * cos(theta) * t1 - 1/2 * g * t1^2 and 3h = u * cos(theta) * t - 1/2 * g * t^2.
- By solving the above equations, we can find the value of t.
Final Answer:
- The horizontal velocity of the stone is 12 m/s. (Option c)
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A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1?
Question Description
A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1?.
A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1?.
Solutions for A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s−1 and the stone still hits the bird while descending, then the horizontal velocity of stone isa. 8 m s−1b. 15 m s−1c. 12 m s−1d. 9 m s−1? in English & in Hindi are available as part of our courses for UPSC.
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