1. Tardigrade
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  4. A ball of mass 1 kg is projected with a velocity of 20 √2 m / s from the origin of an xy co-ordinate axis system at an angle 45° with x-axis (horizontal). The angular momentum [In SI units] of the ball about the point of projection after 2 s of projection is [take .g=10 m / s 2] ( y-axis is taken as vertical)

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Q. A ball of mass $1\, kg$ is projected with a velocity of $20\, \sqrt{2} m / s$ from the origin of an $xy$ co-ordinate axis system at an angle $45^{\circ}$ with $x$-axis (horizontal). The angular momentum [In SI units] of the ball about the point of projection after $2\, s$ of projection is [take $\left.g=10\, m / s ^{2}\right]$ ( $y$-axis is taken as vertical)

System of Particles and Rotational Motion

Solution:

Time of flight $T=\frac{2 u \sin \theta}{g}=\frac{2(20 \sqrt{2}) \frac{1}{\sqrt{2}}}{10}=4$ second
$\Rightarrow $ After 2 second particle will be at maximum height of the projectile $L=m v r_{\perp}$
$r_{\perp}=H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}=20\, m$
So $L=(1)(20)(20)=400(-\hat{k})$