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A particle is projected horizontally wit...

A particle is projected horizontally with speed u from the top of a plane inclined at an angle `theta` with the horizontal. How far from the point of projection will the particle strike the plane?

Text Solution

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`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)/g tan theta`
`x=ut =(2u^(2))/g tan theta, therefore R=(2u^(2))/g tan theta sec theta`
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