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The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .?
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The equation of projectile trajectory of a projectile thrown from a le...
To find the answers to the given options, let's analyze the given equation and its properties.
1. Equation of Projectile Trajectory:
The equation of the projectile trajectory is given by y = ax - bx^2, where y represents the vertical distance and x represents the horizontal distance.
2. Range of the Projectile:
The range of a projectile is the horizontal distance covered by it before hitting the ground. To find the range, we need to determine the value of x when y = 0.
0 = ax - bx^2
bx^2 = ax
x^2 = a/b
x = √(a/b)
Therefore, the range of the projectile is √(a/b), which confirms option A.
3. Velocity of the Projectile:
To find the velocity of the projectile at x = a/2b, we need to differentiate the equation of the trajectory with respect to time (t).
dy/dt = a - 2bx(dx/dt)
At the maximum height of the projectile, dy/dt = 0 since the vertical velocity becomes zero momentarily. Therefore, we have:
0 = a - 2bx(dx/dt)
dx/dt = a/2b
So, the velocity of the projectile at x = a/2b is a/2b, which does not become zero. Hence, option B is incorrect.
4. Maximum Height of the Projectile:
The maximum height attained by the projectile is the highest point on the trajectory. To find this, we need to determine the value of y when dy/dx = 0.
dy/dx = a - 2bx
0 = a - 2bx
2bx = a
x = a/2b
Substituting this value of x into the equation of the trajectory:
y = a(a/2b) - b(a/2b)^2
y = a^2/2b - a^2/4b
y = a^2/4b
Therefore, the maximum height attained by the projectile is a^2/4b, confirming option C.
5. Angle of the Projectile:
The angle of the projectile can be found using the equation:
tan(θ) = (dy/dx)
Differentiating the equation of the trajectory with respect to x:
dy/dx = a - 2bx
tan(θ) = a - 2bx
Therefore, the angle of the projectile is tan^(-1)(a), confirming option D.
In conclusion, options A, C, and D are correct based on the analysis of the given equation and its properties.
1. Equation of Projectile Trajectory:
The equation of the projectile trajectory is given by y = ax - bx^2, where y represents the vertical distance and x represents the horizontal distance.
2. Range of the Projectile:
The range of a projectile is the horizontal distance covered by it before hitting the ground. To find the range, we need to determine the value of x when y = 0.
0 = ax - bx^2
bx^2 = ax
x^2 = a/b
x = √(a/b)
Therefore, the range of the projectile is √(a/b), which confirms option A.
3. Velocity of the Projectile:
To find the velocity of the projectile at x = a/2b, we need to differentiate the equation of the trajectory with respect to time (t).
dy/dt = a - 2bx(dx/dt)
At the maximum height of the projectile, dy/dt = 0 since the vertical velocity becomes zero momentarily. Therefore, we have:
0 = a - 2bx(dx/dt)
dx/dt = a/2b
So, the velocity of the projectile at x = a/2b is a/2b, which does not become zero. Hence, option B is incorrect.
4. Maximum Height of the Projectile:
The maximum height attained by the projectile is the highest point on the trajectory. To find this, we need to determine the value of y when dy/dx = 0.
dy/dx = a - 2bx
0 = a - 2bx
2bx = a
x = a/2b
Substituting this value of x into the equation of the trajectory:
y = a(a/2b) - b(a/2b)^2
y = a^2/2b - a^2/4b
y = a^2/4b
Therefore, the maximum height attained by the projectile is a^2/4b, confirming option C.
5. Angle of the Projectile:
The angle of the projectile can be found using the equation:
tan(θ) = (dy/dx)
Differentiating the equation of the trajectory with respect to x:
dy/dx = a - 2bx
tan(θ) = a - 2bx
Therefore, the angle of the projectile is tan^(-1)(a), confirming option D.
In conclusion, options A, C, and D are correct based on the analysis of the given equation and its properties.
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The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .?
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The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .?.
The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of projectile trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax - bx2 , with y - axis in vertical direction and x - axis inn horizontal direction a and b are constant then, A. The range of the projectile is a/b. B. At x = a/2b , the velocity of projectile becomes zero. C. The maximum height artained by projectile is a2/4b. D. The angle of projectile is tan-1 a Ans : A C and D Can you explain .?.
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