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The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)
- a)25°
- b)30°
- c)45°
- d)60°
Correct answer is option 'B'. Can you explain this answer?
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Given:
Height of ceiling, h = 40 m
Initial speed of the ball, u = 56 m/s
Acceleration due to gravity, g = 9.8 m/s²
To find:
Maximum horizontal distance that the ball can travel without hitting the ceiling, at the maximum angle of projection.
Solution:
Let θ be the angle of projection with respect to the horizontal.
The horizontal and vertical components of the initial velocity are:
u_x = u cos θ
u_y = u sin θ
Time taken by the ball to reach the maximum height can be calculated as:
u_y = u sin θ
v_y = 0 (at the maximum height)
g = 9.8 m/s²
Using the equation of motion: v_y = u_y - gt
0 = u sin θ - gt
t = u sin θ / g
The maximum height reached by the ball can be calculated as:
h = v_yt - 1/2 gt²
h = 0 - 1/2 g (u sin θ / g)²
h = (u sin θ)² / (2g)
The horizontal distance travelled by the ball can be calculated as:
d = u_x t
d = u cos θ (u sin θ / g)
d = u² sin 2θ / g
As the ball should not hit the ceiling, the maximum height reached by the ball should be less than the height of the ceiling:
(u sin θ)² / (2g) < />
(u sin θ)² < />
sin² θ < 2h/u²="" />
sin θ < √(2h/u²="" />
Substituting the given values, we get:
sin θ < />
sin θ < />
Taking the inverse sine on both sides, we get:
θ < />
Therefore, the maximum angle of projection for the ball to travel the maximum horizontal distance without hitting the ceiling is 27.6°, which is closest to option B (30°).
Height of ceiling, h = 40 m
Initial speed of the ball, u = 56 m/s
Acceleration due to gravity, g = 9.8 m/s²
To find:
Maximum horizontal distance that the ball can travel without hitting the ceiling, at the maximum angle of projection.
Solution:
Let θ be the angle of projection with respect to the horizontal.
The horizontal and vertical components of the initial velocity are:
u_x = u cos θ
u_y = u sin θ
Time taken by the ball to reach the maximum height can be calculated as:
u_y = u sin θ
v_y = 0 (at the maximum height)
g = 9.8 m/s²
Using the equation of motion: v_y = u_y - gt
0 = u sin θ - gt
t = u sin θ / g
The maximum height reached by the ball can be calculated as:
h = v_yt - 1/2 gt²
h = 0 - 1/2 g (u sin θ / g)²
h = (u sin θ)² / (2g)
The horizontal distance travelled by the ball can be calculated as:
d = u_x t
d = u cos θ (u sin θ / g)
d = u² sin 2θ / g
As the ball should not hit the ceiling, the maximum height reached by the ball should be less than the height of the ceiling:
(u sin θ)² / (2g) < />
(u sin θ)² < />
sin² θ < 2h/u²="" />
sin θ < √(2h/u²="" />
Substituting the given values, we get:
sin θ < />
sin θ < />
Taking the inverse sine on both sides, we get:
θ < />
Therefore, the maximum angle of projection for the ball to travel the maximum horizontal distance without hitting the ceiling is 27.6°, which is closest to option B (30°).
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The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. Can you explain this answer?
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The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. 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 The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. 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 The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. Can you explain this answer?.
The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. 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 The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. 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 The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is (Take g = 9.8 ms-2)a)25°b)30°c)45°d)60°Correct answer is option 'B'. Can you explain this answer?.
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