question_answer

A particle of mass 1 kg is projected from ground with velocity \[20\text{ }m{{s}^{-1}}\]at \[{{45}^{o}}\] from horizontal. Magnitude of its angular momentum just before it hits the ground about the point of projection will be \[\left( g=10m{{s}^{-2}} \right)\]

A)  \[40\sqrt{2}\frac{kg{{m}^{2}}}{\sec }\] 

B)  \[400\sqrt{2}\frac{kg{{m}^{2}}}{\sec }\]

C)  \[800\frac{kg{{m}^{2}}}{\sec }\]                              

D)  None

Correct Answer: B

Solution :

\[R=\frac{{{u}^{2}}\sin 2\theta }{g}=40\] \[{{L}_{0}}=\] line of linear moment. shortest distance \[=(mv\sin \theta )\times R\]