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The ball is projected with 20 root 2 metre per second at an angle 45°horizontal the angular velocity of the particle at highest point of its journey about point of projection is

Ramani , 7 Years ago
Grade 11
anser 1 Answers
Khimraj
 

max height ,H= v2sin2θ/2g

velocity at max height = ux= vcosθ

angular velocity = uxH

and putting θ =450

angular velocity = v3/g4√2

Hope it clears.

Last Activity: 7 Years ago
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