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A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :
- a)4.4 m
- b)2.4 m
- c)3.6 m
- d)1.6 m
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A ball projected from ground at an angle of 45° just clears a wall in...
As ball is projected at an angle 45° to the horizontal therefore Range = 4H
or 10 = 4H ⇒ H = 10/4 = 2.5 m
(∵ Range = 4 m + 6 m = 10m)
Maximum height, H = u2sin2θ
Or, u = √100 = 10 ms-1
Height of wall PA
= OA tan θ - 1/2 g(OA)2/2u2cos2 θ
Most Upvoted Answer
A ball projected from ground at an angle of 45° just clears a wall in...
To solve this problem, let's break it down into different components and analyze them step by step.
Given:
- Angle of projection: 45°
- Distance of the point of projection from the foot of the wall: 4 m
- Distance of the ball striking the ground on the other side of the wall: 6 m
We need to find the height of the wall.
1. Initial velocity:
When a projectile is launched at an angle, its initial velocity can be resolved into horizontal and vertical components. Since the angle of projection is 45°, the horizontal and vertical components are equal.
Let's assume the initial velocity of the ball is 'v'. Therefore, the horizontal component of velocity is v cos(45°) and the vertical component of velocity is v sin(45°).
2. Time of flight:
The time of flight is the total time taken by the ball to reach the ground after being projected. Since the vertical motion is symmetric, the time taken to reach the maximum height is equal to the time taken to reach the ground.
Let's assume the time of flight is 't'.
3. Horizontal distance:
The horizontal distance covered by the ball can be calculated using the horizontal component of velocity and the time of flight.
Horizontal distance = Horizontal component of velocity × Time of flight
Horizontal distance = v cos(45°) × t
4. Vertical distance:
The vertical distance covered by the ball can be calculated using the vertical component of velocity and the time of flight.
Vertical distance = Vertical component of velocity × Time of flight - (0.5 × g × t^2)
Vertical distance = v sin(45°) × t - (0.5 × g × t^2)
where 'g' is the acceleration due to gravity (approximately 9.8 m/s^2).
5. Height of the wall:
The height of the wall can be determined by subtracting the vertical distance covered by the ball from the initial height of projection.
Height of the wall = Vertical distance - Initial height of projection
Now, let's substitute the given values and solve the equations:
Horizontal distance = (v cos(45°)) × t = 6 m
Vertical distance = (v sin(45°)) × t - (0.5 × g × t^2)
Height of the wall = Vertical distance - Initial height of projection
We know that the horizontal and vertical components of velocity are equal for a 45° angle of projection, so we can simplify the equations further:
Horizontal distance = (v cos(45°)) × t = 6 m
Vertical distance = (v sin(45°)) × t - (0.5 × g × t^2)
Height of the wall = Vertical distance - 4 m
Since the ball just clears the wall, the height of the wall is equal to the vertical distance covered by the ball:
Height of the wall = Vertical distance - 4 m = 6 m - 4 m = 2 m
Therefore, the height of the wall is 2 m. Hence, the correct answer is option 'B'.
Given:
- Angle of projection: 45°
- Distance of the point of projection from the foot of the wall: 4 m
- Distance of the ball striking the ground on the other side of the wall: 6 m
We need to find the height of the wall.
1. Initial velocity:
When a projectile is launched at an angle, its initial velocity can be resolved into horizontal and vertical components. Since the angle of projection is 45°, the horizontal and vertical components are equal.
Let's assume the initial velocity of the ball is 'v'. Therefore, the horizontal component of velocity is v cos(45°) and the vertical component of velocity is v sin(45°).
2. Time of flight:
The time of flight is the total time taken by the ball to reach the ground after being projected. Since the vertical motion is symmetric, the time taken to reach the maximum height is equal to the time taken to reach the ground.
Let's assume the time of flight is 't'.
3. Horizontal distance:
The horizontal distance covered by the ball can be calculated using the horizontal component of velocity and the time of flight.
Horizontal distance = Horizontal component of velocity × Time of flight
Horizontal distance = v cos(45°) × t
4. Vertical distance:
The vertical distance covered by the ball can be calculated using the vertical component of velocity and the time of flight.
Vertical distance = Vertical component of velocity × Time of flight - (0.5 × g × t^2)
Vertical distance = v sin(45°) × t - (0.5 × g × t^2)
where 'g' is the acceleration due to gravity (approximately 9.8 m/s^2).
5. Height of the wall:
The height of the wall can be determined by subtracting the vertical distance covered by the ball from the initial height of projection.
Height of the wall = Vertical distance - Initial height of projection
Now, let's substitute the given values and solve the equations:
Horizontal distance = (v cos(45°)) × t = 6 m
Vertical distance = (v sin(45°)) × t - (0.5 × g × t^2)
Height of the wall = Vertical distance - Initial height of projection
We know that the horizontal and vertical components of velocity are equal for a 45° angle of projection, so we can simplify the equations further:
Horizontal distance = (v cos(45°)) × t = 6 m
Vertical distance = (v sin(45°)) × t - (0.5 × g × t^2)
Height of the wall = Vertical distance - 4 m
Since the ball just clears the wall, the height of the wall is equal to the vertical distance covered by the ball:
Height of the wall = Vertical distance - 4 m = 6 m - 4 m = 2 m
Therefore, the height of the wall is 2 m. Hence, the correct answer is option 'B'.
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A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. Can you explain this answer?
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A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. 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 A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. Can you explain this answer?.
A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. 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 A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :a)4.4 mb)2.4 mc)3.6 md)1.6 mCorrect answer is option 'B'. Can you explain this answer?.
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