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A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 9√3 m/s at the same instant as the shell is fired, then find u/25 (in m/s).
Take g = 10 m/s2.
Take g = 10 m/s2.
Correct answer is '2'. Can you explain this answer?
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A shell is fired from a point O at an angle of 60o with a speed of 40 ...
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A shell is fired from a point O at an angle of 60o with a speed of 40 ...
To solve this problem, we'll break it down into different parts and analyze each part separately.
**Part 1: Shell's trajectory and impact point**
1. The shell is fired from point O at an angle of 60 degrees with a speed of 40 m/s.
2. We can break the initial velocity of the shell into its horizontal and vertical components. The horizontal component of velocity (Vx) remains constant throughout the motion, while the vertical component of velocity (Vy) changes due to the effect of gravity.
3. The initial vertical component of velocity (Vy) can be calculated using the formula Vy = V * sin(θ), where V is the initial speed of the shell and θ is the angle of elevation.
Therefore, Vy = 40 * sin(60) = 40 * (√3/2) = 20√3 m/s.
4. The time taken by the shell to reach the horizontal plane is the same as the time taken to reach the maximum height since the vertical motion is symmetrical.
5. The time of flight (T) can be calculated using the formula T = 2 * Vy / g, where g is the acceleration due to gravity.
Therefore, T = 2 * (20√3) / 10 = 4√3 s.
6. The horizontal distance traveled by the shell can be calculated using the formula Dx = Vx * T, where Vx is the horizontal component of velocity and T is the time of flight.
The horizontal component of velocity (Vx) can be calculated using the formula Vx = V * cos(θ), where V is the initial speed of the shell and θ is the angle of elevation.
Therefore, Vx = 40 * cos(60) = 40 * (1/2) = 20 m/s.
Therefore, Dx = 20 * (4√3) = 80√3 m.
7. The shell strikes the horizontal plane at a point A, which is 80√3 meters away from the point of firing.
**Part 2: Target's motion**
1. The target starts to rise vertically from point A at the same instant the shell is fired.
2. The target rises with a constant speed of 93 m/s.
3. We need to find the speed (u) at which the gun is fired for the second time to hit the target.
4. Let's assume the time taken by the second shell to hit the target is T2.
5. The horizontal distance traveled by the second shell can be calculated using the formula Dx = u * T2, where Dx is the horizontal distance and u is the speed of the second shell.
6. The vertical distance traveled by the target can be calculated using the formula Dy = 93 * T2, where Dy is the vertical distance traveled by the target.
7. The horizontal distance traveled by the target is the same as the horizontal distance traveled by the second shell since they both hit the target at the same point.
Therefore, Dx = 80√3 m.
8. The vertical distance traveled by the target can also be calculated using the formula Dy = Vyt + (1/2) * gt^2, where Vyt is the initial vertical component of velocity of the target and g is the acceleration due to gravity.
Since the target starts from rest vertically
**Part 1: Shell's trajectory and impact point**
1. The shell is fired from point O at an angle of 60 degrees with a speed of 40 m/s.
2. We can break the initial velocity of the shell into its horizontal and vertical components. The horizontal component of velocity (Vx) remains constant throughout the motion, while the vertical component of velocity (Vy) changes due to the effect of gravity.
3. The initial vertical component of velocity (Vy) can be calculated using the formula Vy = V * sin(θ), where V is the initial speed of the shell and θ is the angle of elevation.
Therefore, Vy = 40 * sin(60) = 40 * (√3/2) = 20√3 m/s.
4. The time taken by the shell to reach the horizontal plane is the same as the time taken to reach the maximum height since the vertical motion is symmetrical.
5. The time of flight (T) can be calculated using the formula T = 2 * Vy / g, where g is the acceleration due to gravity.
Therefore, T = 2 * (20√3) / 10 = 4√3 s.
6. The horizontal distance traveled by the shell can be calculated using the formula Dx = Vx * T, where Vx is the horizontal component of velocity and T is the time of flight.
The horizontal component of velocity (Vx) can be calculated using the formula Vx = V * cos(θ), where V is the initial speed of the shell and θ is the angle of elevation.
Therefore, Vx = 40 * cos(60) = 40 * (1/2) = 20 m/s.
Therefore, Dx = 20 * (4√3) = 80√3 m.
7. The shell strikes the horizontal plane at a point A, which is 80√3 meters away from the point of firing.
**Part 2: Target's motion**
1. The target starts to rise vertically from point A at the same instant the shell is fired.
2. The target rises with a constant speed of 93 m/s.
3. We need to find the speed (u) at which the gun is fired for the second time to hit the target.
4. Let's assume the time taken by the second shell to hit the target is T2.
5. The horizontal distance traveled by the second shell can be calculated using the formula Dx = u * T2, where Dx is the horizontal distance and u is the speed of the second shell.
6. The vertical distance traveled by the target can be calculated using the formula Dy = 93 * T2, where Dy is the vertical distance traveled by the target.
7. The horizontal distance traveled by the target is the same as the horizontal distance traveled by the second shell since they both hit the target at the same point.
Therefore, Dx = 80√3 m.
8. The vertical distance traveled by the target can also be calculated using the formula Dy = Vyt + (1/2) * gt^2, where Vyt is the initial vertical component of velocity of the target and g is the acceleration due to gravity.
Since the target starts from rest vertically
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A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer?
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A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer?.
A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer?.
Solutions for A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice A shell is fired from a point O at an angle of 60o with a speed of 40 m/s and it strikes a horizontal plane through O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed u. If it hits the target which starts to rise vertically from A with a constant speed 93 m/s at the same instant as the shell is fired, then find u/25(in m/s).Take g = 10 m/s2. Correct answer is '2'. Can you explain this answer? tests, examples and also practice JEE tests.
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