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Grade: 12th pass
        A particle of mass 1 kg is projected with velocity 20√2m/s. at45° with ground when the particle is at a highest point angular velocity about point of projection
2 years ago

Answers : (1)

jitender
114 Points
							V at high as point V = 20√2cos45= 20m/sDisplacement from initial point to highest point D = √(H^2+(R/2)^2) Where H =[( 20√2) ^2 sin^2 45]/2g = 40mAnd R=[( 20√2) ^2 sin 2×45]/g = 80mNow D = 40√2mAngular velocity ₩ = V_/DV_= velocity perpendicular to DV_ = Vcos 45 = component of V perpendicular to D₩ = (20×1/√2)/40√2 = 1/4 = 0.25rad/s
						
2 years ago
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