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The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be
- a)R + H/2
- b)R +H
- c)R + 3H/2
- d)R + 2H
Correct answer is option 'D'. Can you explain this answer?
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Solution:
Given, horizontal range of projectile = R and maximum height attained by it = H.
Let the time taken to reach maximum height H be t.
Now, using the equation of motion, s = ut + 1/2 at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.
- Initial vertical velocity, uy = usinθ
- Vertical acceleration, ay = -g
- Displacement in y-direction, sy = H
- Time taken to reach maximum height, t = uy/ay = usinθ/g
Now, using the equation of motion, v = u + at, where v is the final velocity.
- Final vertical velocity, vy = 0
- Using vy = uy + ayt, we get:
- t = 2usinθ/g
Using the equation of motion, s = ut + 1/2 at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.
- Initial horizontal velocity, ux = ucosθ
- Horizontal acceleration, ax = g/2
- Displacement in x-direction, sx = R
- Time taken, t = 2usinθ/g
Using the equation of motion, s = ut + 1/2 at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.
- Horizontal displacement, sx = uxt + 1/2 axt^2
- Substituting the values, we get:
- R = ucosθ × 2usinθ/g + 1/2 g/2 (2usinθ/g)^2
- R = u^2 sin 2θ/g + u^2 sin^2θ/g
- R = u^2/g (sin 2θ + sin^2θ)
- Using the identity, sin 2θ + sin^2θ = (3/2)sin^2θ + (1/2), we get:
- R = u^2/g [(3/2)sin^2θ + (1/2)]
- R = 3/2 (u^2/g) sin^2θ + 1/2 (u^
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The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer?
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The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer?.
The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now bea)R + H/2b)R +Hc)R + 3H/2d)R + 2HCorrect answer is option 'D'. Can you explain this answer?.
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