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A ball is thrown from a point with a speed v0 at an angle fo projection 
. From the same point and at the same instant, a person starts running with a constant speed 
to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?
. From the same point and at the same instant, a person starts running with a constant speed to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]
- a)Yes, 60°
- b)Yes, 30°
- c)No
- d)Yes, 45°
Correct answer is option 'A'. Can you explain this answer?
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A ball is thrown from a point with a speed v0at an angle fo projection...
Man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man ie.
vo cos θ = vo/2
or Cos θ = 1/2
or θ = 60o
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A ball is thrown from a point with a speed v0at an angle fo projection...
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A ball is thrown from a point with a speed v0at an angle fo projection...
Analysis:
The key to solving this problem lies in understanding the relationship between the speed of the ball and the speed of the person running.
Relative Motion:
When the person starts running with a speed of v0/2, the effective speed of the person relative to the ball becomes v0 + v0/2 = 3v0/2.
Horizontal and Vertical Components:
The horizontal component of the ball's velocity remains v0cosθ and the vertical component remains v0sinθ.
Time of Flight:
The time taken by the ball to reach the maximum height is given by t = v0sinθ/g and the total time of flight is 2t.
Distance covered by the person:
The distance covered by the person running is (v0/2) * 2t = v0t = v0(v0sinθ/g).
Condition for Catching the Ball:
For the person to catch the ball, the distance covered by the person running should be equal to the horizontal distance covered by the ball.
Solving for θ:
By equating the two distances, we get v0(v0sinθ/g) = v0cosθ * 2t.
Solving further, we get sinθ = 2/3, which corresponds to θ = 60°.
Therefore, the person will be able to catch the ball if it is thrown at an angle of 60°.
The key to solving this problem lies in understanding the relationship between the speed of the ball and the speed of the person running.
Relative Motion:
When the person starts running with a speed of v0/2, the effective speed of the person relative to the ball becomes v0 + v0/2 = 3v0/2.
Horizontal and Vertical Components:
The horizontal component of the ball's velocity remains v0cosθ and the vertical component remains v0sinθ.
Time of Flight:
The time taken by the ball to reach the maximum height is given by t = v0sinθ/g and the total time of flight is 2t.
Distance covered by the person:
The distance covered by the person running is (v0/2) * 2t = v0t = v0(v0sinθ/g).
Condition for Catching the Ball:
For the person to catch the ball, the distance covered by the person running should be equal to the horizontal distance covered by the ball.
Solving for θ:
By equating the two distances, we get v0(v0sinθ/g) = v0cosθ * 2t.
Solving further, we get sinθ = 2/3, which corresponds to θ = 60°.
Therefore, the person will be able to catch the ball if it is thrown at an angle of 60°.
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A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer?
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A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer?.
A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown from a point with a speed v0at an angle fo projection. From the same point and at the same instant, a person starts running with a constant speedto catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection ?[AIEEE 2004]a)Yes, 60°b)Yes, 30°c)Nod)Yes, 45°Correct answer is option 'A'. Can you explain this answer?.
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