InterviewSolution
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A particle is projected with a velocity 10 m//s at an angle `37^(@)` to the horizontal. Find the location at which the particle is at a height 1 m from point of projection. |
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Answer» Correct Answer - 1.6 m, 8 m. `y=x tan theta - (gx^(2))/(2u^(2) cos^(2) theta)` for `y=1, theta=37^(@), u=10 m//s` `1=3/4x - (10x^(2))/(2xx100xx((16),(25)))` `1=3/4x -5/64 x^(2)` `5x^(2)-48x+64 =0` `5x^(2)-40x-8x+64=0` `x=8m, 1.6 m` |
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