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A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is `vecv=(6hati+2hatj)m//s` (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is `(g=10m//s^(2))`A. `45^(@)`B. `60^(@)`C. `30^(@)`D. `tan^(-1)(3//4)` |
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Answer» Correct Answer - C `vecV= 6hati+2hatj" " h =0.4m` Given `V_(x)=6=u_(x)` (in projectile) `V_(y)=2=sqrt(u_(y)^(2)-2gh)` `therefore U_(y)^(2)=4+2gh=4+2(10)/(0.4)=12` `therefore U_(y) =sqrt(12) ms^(-1)` angle of projection ` tan theta=(U_(y))/(U_(x))=(sqrt(12))/(3)=(1)/(sqrt(3))` `therefore theta= 30^(@)` |
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