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A particle is projected from a horizontal plane with a velocity of 8√2 metre per second at some angle. if its velocity at the maximum point is find to be 8 metre per second. find its range |
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Answer» The initial velocity is v =8√2 m/s. and we know that Velocity at max. point is Vcosθ so Vcosθ = 8 => 8√2cosθ = 8 => cosθ = 8/8√2 = 1/√2 therefore θ = 45° now. range is u²sin2θ/g = (8√2)²sin(2*45)/10 = 8*8*2/10 = 128/10 = 12.8m |
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