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Motion in Plane

269976 Equations of motion of a projectile are given by \(x=36 t\) and \(2 y=96 t-98 t^{2} m\). The angle of projection is equal to \(\quad(2011-M)\)

1 \(\sin ^{-1} \emptyset^{-3} \theta\)
2 \(\sin ^{-1} \square^{-4} \square\)
3 \(\sin ^{-1} \square^{-4} \square\)
4 \(\sin ^{-1} \square \frac{3}{5} \theta\)
Motion in Plane

269977 \(\bar{A}\) and \(\bar{B}\) are twovectors of equal magnitude and ' \(\theta\) ' is the angle between them. The angle between \(\bar{A}\) or \(\bar{B}\) with their resultant is( \((\mathrm{E}-2010)\)

1 \(\frac{\theta}{4}\)
2 \(\frac{\theta}{2}\)
3 \(2 \theta\)
4 0
Motion in Plane

269978 If a body is projected with an angle \(\theta\) to the horizontal then
\((\mathrm{E}-2008)\)

1 it's velocity always perpendicular to its acceleration
2 its velocity becomes zero at its maximum height
3 it's velocity makes zero angle with the horizontal at its maximum height
4 the body just before hitting the ground, the direction of velocity coincides with the acceleration.
Motion in Plane

269979 A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight then the value of maximum range is (acceleration due to gravity \(=\mathbf{g})(2014\) - E)

1 \(\frac{g^{2} T}{2}\)
2 \(\frac{g T}{2}\)
3 \(\frac{g T^{2}}{2}\)
4 \(\frac{g^{2} T^{2}}{2}\)
Motion in Plane

269980 The path of projectile is given by the equation \(y=a x-b x^{2}\), where ' \(\mathbf{a}\) ' and ' \(b\) ' are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )

1 \(\frac{2 a^{2}}{b}, \operatorname{Tan}^{-1}(a)\)
2 \(\frac{b^{2}}{2 a}, \operatorname{Tan}^{-1}(b)\)
3 \(\frac{a^{2}}{b}, \operatorname{Tan}^{-1}(2 b)\)
4 \(\frac{a^{2}}{4 b}, \operatorname{Tan}^{-1}(a)\)
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Motion in Plane

269976 Equations of motion of a projectile are given by \(x=36 t\) and \(2 y=96 t-98 t^{2} m\). The angle of projection is equal to \(\quad(2011-M)\)

1 \(\sin ^{-1} \emptyset^{-3} \theta\)
2 \(\sin ^{-1} \square^{-4} \square\)
3 \(\sin ^{-1} \square^{-4} \square\)
4 \(\sin ^{-1} \square \frac{3}{5} \theta\)
Motion in Plane

269977 \(\bar{A}\) and \(\bar{B}\) are twovectors of equal magnitude and ' \(\theta\) ' is the angle between them. The angle between \(\bar{A}\) or \(\bar{B}\) with their resultant is( \((\mathrm{E}-2010)\)

1 \(\frac{\theta}{4}\)
2 \(\frac{\theta}{2}\)
3 \(2 \theta\)
4 0
Motion in Plane

269978 If a body is projected with an angle \(\theta\) to the horizontal then
\((\mathrm{E}-2008)\)

1 it's velocity always perpendicular to its acceleration
2 its velocity becomes zero at its maximum height
3 it's velocity makes zero angle with the horizontal at its maximum height
4 the body just before hitting the ground, the direction of velocity coincides with the acceleration.
Motion in Plane

269979 A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight then the value of maximum range is (acceleration due to gravity \(=\mathbf{g})(2014\) - E)

1 \(\frac{g^{2} T}{2}\)
2 \(\frac{g T}{2}\)
3 \(\frac{g T^{2}}{2}\)
4 \(\frac{g^{2} T^{2}}{2}\)
Motion in Plane

269980 The path of projectile is given by the equation \(y=a x-b x^{2}\), where ' \(\mathbf{a}\) ' and ' \(b\) ' are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )

1 \(\frac{2 a^{2}}{b}, \operatorname{Tan}^{-1}(a)\)
2 \(\frac{b^{2}}{2 a}, \operatorname{Tan}^{-1}(b)\)
3 \(\frac{a^{2}}{b}, \operatorname{Tan}^{-1}(2 b)\)
4 \(\frac{a^{2}}{4 b}, \operatorname{Tan}^{-1}(a)\)
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Motion in Plane

269976 Equations of motion of a projectile are given by \(x=36 t\) and \(2 y=96 t-98 t^{2} m\). The angle of projection is equal to \(\quad(2011-M)\)

1 \(\sin ^{-1} \emptyset^{-3} \theta\)
2 \(\sin ^{-1} \square^{-4} \square\)
3 \(\sin ^{-1} \square^{-4} \square\)
4 \(\sin ^{-1} \square \frac{3}{5} \theta\)
Motion in Plane

269977 \(\bar{A}\) and \(\bar{B}\) are twovectors of equal magnitude and ' \(\theta\) ' is the angle between them. The angle between \(\bar{A}\) or \(\bar{B}\) with their resultant is( \((\mathrm{E}-2010)\)

1 \(\frac{\theta}{4}\)
2 \(\frac{\theta}{2}\)
3 \(2 \theta\)
4 0
Motion in Plane

269978 If a body is projected with an angle \(\theta\) to the horizontal then
\((\mathrm{E}-2008)\)

1 it's velocity always perpendicular to its acceleration
2 its velocity becomes zero at its maximum height
3 it's velocity makes zero angle with the horizontal at its maximum height
4 the body just before hitting the ground, the direction of velocity coincides with the acceleration.
Motion in Plane

269979 A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight then the value of maximum range is (acceleration due to gravity \(=\mathbf{g})(2014\) - E)

1 \(\frac{g^{2} T}{2}\)
2 \(\frac{g T}{2}\)
3 \(\frac{g T^{2}}{2}\)
4 \(\frac{g^{2} T^{2}}{2}\)
Motion in Plane

269980 The path of projectile is given by the equation \(y=a x-b x^{2}\), where ' \(\mathbf{a}\) ' and ' \(b\) ' are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )

1 \(\frac{2 a^{2}}{b}, \operatorname{Tan}^{-1}(a)\)
2 \(\frac{b^{2}}{2 a}, \operatorname{Tan}^{-1}(b)\)
3 \(\frac{a^{2}}{b}, \operatorname{Tan}^{-1}(2 b)\)
4 \(\frac{a^{2}}{4 b}, \operatorname{Tan}^{-1}(a)\)
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Motion in Plane

269976 Equations of motion of a projectile are given by \(x=36 t\) and \(2 y=96 t-98 t^{2} m\). The angle of projection is equal to \(\quad(2011-M)\)

1 \(\sin ^{-1} \emptyset^{-3} \theta\)
2 \(\sin ^{-1} \square^{-4} \square\)
3 \(\sin ^{-1} \square^{-4} \square\)
4 \(\sin ^{-1} \square \frac{3}{5} \theta\)
Motion in Plane

269977 \(\bar{A}\) and \(\bar{B}\) are twovectors of equal magnitude and ' \(\theta\) ' is the angle between them. The angle between \(\bar{A}\) or \(\bar{B}\) with their resultant is( \((\mathrm{E}-2010)\)

1 \(\frac{\theta}{4}\)
2 \(\frac{\theta}{2}\)
3 \(2 \theta\)
4 0
Motion in Plane

269978 If a body is projected with an angle \(\theta\) to the horizontal then
\((\mathrm{E}-2008)\)

1 it's velocity always perpendicular to its acceleration
2 its velocity becomes zero at its maximum height
3 it's velocity makes zero angle with the horizontal at its maximum height
4 the body just before hitting the ground, the direction of velocity coincides with the acceleration.
Motion in Plane

269979 A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight then the value of maximum range is (acceleration due to gravity \(=\mathbf{g})(2014\) - E)

1 \(\frac{g^{2} T}{2}\)
2 \(\frac{g T}{2}\)
3 \(\frac{g T^{2}}{2}\)
4 \(\frac{g^{2} T^{2}}{2}\)
Motion in Plane

269980 The path of projectile is given by the equation \(y=a x-b x^{2}\), where ' \(\mathbf{a}\) ' and ' \(b\) ' are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )

1 \(\frac{2 a^{2}}{b}, \operatorname{Tan}^{-1}(a)\)
2 \(\frac{b^{2}}{2 a}, \operatorname{Tan}^{-1}(b)\)
3 \(\frac{a^{2}}{b}, \operatorname{Tan}^{-1}(2 b)\)
4 \(\frac{a^{2}}{4 b}, \operatorname{Tan}^{-1}(a)\)
NEET DIGITAL LIBRARY
Motion in Plane

269976 Equations of motion of a projectile are given by \(x=36 t\) and \(2 y=96 t-98 t^{2} m\). The angle of projection is equal to \(\quad(2011-M)\)

1 \(\sin ^{-1} \emptyset^{-3} \theta\)
2 \(\sin ^{-1} \square^{-4} \square\)
3 \(\sin ^{-1} \square^{-4} \square\)
4 \(\sin ^{-1} \square \frac{3}{5} \theta\)
Motion in Plane

269977 \(\bar{A}\) and \(\bar{B}\) are twovectors of equal magnitude and ' \(\theta\) ' is the angle between them. The angle between \(\bar{A}\) or \(\bar{B}\) with their resultant is( \((\mathrm{E}-2010)\)

1 \(\frac{\theta}{4}\)
2 \(\frac{\theta}{2}\)
3 \(2 \theta\)
4 0
Motion in Plane

269978 If a body is projected with an angle \(\theta\) to the horizontal then
\((\mathrm{E}-2008)\)

1 it's velocity always perpendicular to its acceleration
2 its velocity becomes zero at its maximum height
3 it's velocity makes zero angle with the horizontal at its maximum height
4 the body just before hitting the ground, the direction of velocity coincides with the acceleration.
Motion in Plane

269979 A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight then the value of maximum range is (acceleration due to gravity \(=\mathbf{g})(2014\) - E)

1 \(\frac{g^{2} T}{2}\)
2 \(\frac{g T}{2}\)
3 \(\frac{g T^{2}}{2}\)
4 \(\frac{g^{2} T^{2}}{2}\)
Motion in Plane

269980 The path of projectile is given by the equation \(y=a x-b x^{2}\), where ' \(\mathbf{a}\) ' and ' \(b\) ' are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )

1 \(\frac{2 a^{2}}{b}, \operatorname{Tan}^{-1}(a)\)
2 \(\frac{b^{2}}{2 a}, \operatorname{Tan}^{-1}(b)\)
3 \(\frac{a^{2}}{b}, \operatorname{Tan}^{-1}(2 b)\)
4 \(\frac{a^{2}}{4 b}, \operatorname{Tan}^{-1}(a)\)