$A$ 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):
- AZero
- B$mv^3 / (4\sqrt{2}g)$
- C$mv^3 / (\sqrt{2}g)$
- D$mv^2 / 2g$
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$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:
$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)
Difficult
View SolutionThe ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is
Medium
View SolutionThe angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)
$A$ 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 $\,m$)? $(g = 10 \,ms^{-2})$
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