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A particle is projected horizontally with a speed ''u'' rom the top of plane incline at an angle theta with the horizontal. How far from the point of projection will the particle strike the plane ? |
Answer» Solution : `R=sqrt(X^(2) + y^(2)) (y/x = TAN theta)` `=sqrt(x^(2) + (x tan theta)^(2)) =xsqrt(1 + tan^(2) theta) =x sec theta` `x=ut, y=1/2 "gt"^(2), y/x = 1/2("gt"^(2))/(ut)` `tan theta =("gt")/(2u), t=(2u)/gtan theta` `x=ut =(2u^(2))/g tan theta, therefore R=(2u^(2))/g. tan theta.sec theta` |
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