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A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole?
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A vertical pole has a red mark at some height. A stone is projected fr...
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A vertical pole has a red mark at some height. A stone is projected fr...
Given:
- The stone is projected from a fixed point on the ground.
- The stone hits the pole orthogonally, which means it hits the pole at a right angle.
- The red mark on the pole is at a certain height.
- When projected at an angle of 45 degrees, the stone hits the pole 1 m above the mark.
- When projected at an angle of tan(P) = 0.75, the stone hits the pole 1.5 m below the mark.
To find:
- The velocity and angle of projection at which the stone hits the mark orthogonally.
Solution:
Let's break down the problem into smaller parts:
1. Finding the initial velocity of the stone:
When the stone is projected at an angle of 45 degrees, it hits the pole 1 m above the mark. Since it hits the pole orthogonally, the stone travels a distance equal to the height of the mark.
Using the equation of motion:
h = u^2 sin^2(θ) / (2g)
Where:
h = height (1 m)
u = initial velocity
θ = angle of projection (45 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the given values:
1 = u^2 sin^2(45) / (2 * 9.8)
Simplifying the equation:
u^2 * 0.5 = 9.8
u^2 = 9.8 / 0.5
u^2 = 19.6
u = √19.6
u ≈ 4.43 m/s
So, the initial velocity of the stone when projected at 45 degrees is approximately 4.43 m/s.
2. Finding the angle of projection:
When the stone is projected at an angle of tan(P) = 0.75, it hits the pole 1.5 m below the mark. Again, since it hits the pole orthogonally, the stone travels a distance equal to the height of the mark.
Using the equation of motion:
h = u^2 sin^2(θ) / (2g)
Where:
h = height (-1.5 m, as it is below the mark)
u = initial velocity (unknown)
θ = angle of projection (unknown)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the given values:
-1.5 = u^2 sin^2(P) / (2 * 9.8)
Simplifying the equation:
u^2 * sin^2(P) = -1.5 * 2 * 9.8
u^2 * sin^2(0.75) = -29.4
u^2 = -29.4 / sin^2(0.75)
u ≈ √(-29.4 / sin^2(0.75))
Since the square root of a negative number is not defined in real numbers, the given information is not possible. There is no velocity and angle of projection that will make the stone hit the mark orthogonally.
Therefore, the problem has no valid solution.
- The stone is projected from a fixed point on the ground.
- The stone hits the pole orthogonally, which means it hits the pole at a right angle.
- The red mark on the pole is at a certain height.
- When projected at an angle of 45 degrees, the stone hits the pole 1 m above the mark.
- When projected at an angle of tan(P) = 0.75, the stone hits the pole 1.5 m below the mark.
To find:
- The velocity and angle of projection at which the stone hits the mark orthogonally.
Solution:
Let's break down the problem into smaller parts:
1. Finding the initial velocity of the stone:
When the stone is projected at an angle of 45 degrees, it hits the pole 1 m above the mark. Since it hits the pole orthogonally, the stone travels a distance equal to the height of the mark.
Using the equation of motion:
h = u^2 sin^2(θ) / (2g)
Where:
h = height (1 m)
u = initial velocity
θ = angle of projection (45 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the given values:
1 = u^2 sin^2(45) / (2 * 9.8)
Simplifying the equation:
u^2 * 0.5 = 9.8
u^2 = 9.8 / 0.5
u^2 = 19.6
u = √19.6
u ≈ 4.43 m/s
So, the initial velocity of the stone when projected at 45 degrees is approximately 4.43 m/s.
2. Finding the angle of projection:
When the stone is projected at an angle of tan(P) = 0.75, it hits the pole 1.5 m below the mark. Again, since it hits the pole orthogonally, the stone travels a distance equal to the height of the mark.
Using the equation of motion:
h = u^2 sin^2(θ) / (2g)
Where:
h = height (-1.5 m, as it is below the mark)
u = initial velocity (unknown)
θ = angle of projection (unknown)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the given values:
-1.5 = u^2 sin^2(P) / (2 * 9.8)
Simplifying the equation:
u^2 * sin^2(P) = -1.5 * 2 * 9.8
u^2 * sin^2(0.75) = -29.4
u^2 = -29.4 / sin^2(0.75)
u ≈ √(-29.4 / sin^2(0.75))
Since the square root of a negative number is not defined in real numbers, the given information is not possible. There is no velocity and angle of projection that will make the stone hit the mark orthogonally.
Therefore, the problem has no valid solution.
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A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole?
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A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole?.
A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole?.
Solutions for A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? in English & in Hindi are available as part of our courses for JEE.
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A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole?, a detailed solution for A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? has been provided alongside types of A vertical pole has a red mark at some height. A stone is projected from a fixed point on a ground. When projected at an angle of 45′ it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tanP = 0.75, it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole? theory, EduRev gives you an
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