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A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?
a. (400,100)
b . (200,100)
c. (400,200)
d. (200,200)?
a. (400,100)
b . (200,100)
c. (400,200)
d. (200,200)?
Most Upvoted Answer
A projectile is thrown into space so as to have the maximum possible r...
Solution:
Given, maximum possible range = 400m
Let, u be the initial velocity of the projectile and θ be the angle of projection.
The horizontal range of the projectile is given by,
R = u^2 sin 2θ/g
where, g = 9.8 m/s^2 is the acceleration due to gravity.
To get the maximum range, we need to differentiate R w.r.t θ and equate it to zero.
dR/dθ = 2u^2 cos 2θ/g = 0
cos 2θ = 0
2θ = π/2 or 3π/2
θ = π/4 or 3π/4 (taking only first quadrant angles)
Now, the coordinates of the point where the velocity of the projectile is minimum can be obtained by differentiating the x and y components of velocity w.r.t θ and equating them to zero.
Let, vx and vy be the x and y components of velocity respectively.
vx = u cos θ
vy = u sin θ - gt
dvx/dθ = -u sin θ = 0
θ = 0 (minimum value of vx)
dvy/dθ = u cos θ - gt = 0
u cos θ = gt
Substituting θ = π/4, we get
u = gt/cos π/4 = gt/√2
The time of flight of the projectile can be calculated as,
t = 2u sin θ/g = 2u/√2g = 2tmax/√2
where, tmax is the time of flight for θ = π/4.
The minimum velocity of the projectile can be calculated as,
vmin = √(vx^2 + vy^2)
Substituting the values, we get
vmin = u/√2 = 20√2 m/s
The x and y coordinates of the point where the velocity is minimum can be obtained as,
x = u^2 sin 2θ/g * cos θ * t = 200m
y = u^2 sin^2 θ/2g * t = 100m
Therefore, the coordinates of the point where the velocity of the projectile is minimum are (200,100).
Hence, option (b) is the correct answer.
Given, maximum possible range = 400m
Let, u be the initial velocity of the projectile and θ be the angle of projection.
The horizontal range of the projectile is given by,
R = u^2 sin 2θ/g
where, g = 9.8 m/s^2 is the acceleration due to gravity.
To get the maximum range, we need to differentiate R w.r.t θ and equate it to zero.
dR/dθ = 2u^2 cos 2θ/g = 0
cos 2θ = 0
2θ = π/2 or 3π/2
θ = π/4 or 3π/4 (taking only first quadrant angles)
Now, the coordinates of the point where the velocity of the projectile is minimum can be obtained by differentiating the x and y components of velocity w.r.t θ and equating them to zero.
Let, vx and vy be the x and y components of velocity respectively.
vx = u cos θ
vy = u sin θ - gt
dvx/dθ = -u sin θ = 0
θ = 0 (minimum value of vx)
dvy/dθ = u cos θ - gt = 0
u cos θ = gt
Substituting θ = π/4, we get
u = gt/cos π/4 = gt/√2
The time of flight of the projectile can be calculated as,
t = 2u sin θ/g = 2u/√2g = 2tmax/√2
where, tmax is the time of flight for θ = π/4.
The minimum velocity of the projectile can be calculated as,
vmin = √(vx^2 + vy^2)
Substituting the values, we get
vmin = u/√2 = 20√2 m/s
The x and y coordinates of the point where the velocity is minimum can be obtained as,
x = u^2 sin 2θ/g * cos θ * t = 200m
y = u^2 sin^2 θ/2g * t = 100m
Therefore, the coordinates of the point where the velocity of the projectile is minimum are (200,100).
Hence, option (b) is the correct answer.
Community Answer
A projectile is thrown into space so as to have the maximum possible r...
Correct answer is option B
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A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)?
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A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)?.
A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is thrown into space so as to have the maximum possible range equal to 400m.taking the point of projection as the origin,the oordinates of the point where are the velocity of the projectile is minimum are :-?a. (400,100)b . (200,100)c. (400,200)d. (200,200)?.
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