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A particle is projected horizontally wit...

A particle is projected horizontally with a speed `u` from the top of 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 cos 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`
`:. R=(2u^(2))/(g) tan theta sec theta`
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