(B) At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.The momen

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(B) At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.The momentum of the particle at this point is $p = m v_x = mv / \sqrt{2}$.The angular momentum $L$ about the point of projection is given by $L = p \times h$,where $h$ is the maximum height.The maximum height $h$ is given by $h = (v^2 \sin^2 45^{\circ}) / (2g) = (v^2 \cdot (1/2)) / (2g) = v^2 / (4g)$.Substituting these values,we get $L = (mv / \sqrt{2}) \times (v^2 / 4g) = mv^3 / (4\sqrt{2}g)$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Difficult

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g$ is the acceleration due to gravity):

A
Zero
B
$mv^3 / (4\sqrt{2}g)$
C
$mv^3 / (\sqrt{2}g)$
D
$mv^2 / 2g$

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3-2.Motion in Plane

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Horizontal Projectile Motion

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.

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$A$ boy can throw a stone up to a maximum height of $10 \ m$. The maximum horizontal distance that the boy can throw the same stone up to will be .......... $m$.

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Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

MediumAIIMS 2001

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$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

MediumAP EAMCET 2021

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The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The velocity with which the projectile is projected is ......... $m/s$.

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$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g$ is the acceleration due to gravity):

Similar Questions

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.

$A$ boy can throw a stone up to a maximum height of $10 \ m$. The maximum horizontal distance that the boy can throw the same stone up to will be .......... $m$.

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The velocity with which the projectile is projected is ......... $m/s$.

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