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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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The horizontal distance covered in time t,
`y=0+1/2"gt"^2=1/2gxx(x^2)/(u^2)`

Also `y/x=tantheta` or y=x `tantheta" ":.(gx^2)/(2u^2)=xtantheta`
or `x((gx)/(2u^2)-tantheta)=0`
As x=0 is not possible, so x`=(2u^2tantheta)/(g)`
The distance of the point of strike from the point of projection is
`D=sqrt(x^2+y^2)=sqrt(x^2+(xtantheta)^2)`
`=xsqrt(1+tan^2theta)=xsec theta` or
`D=(2u^2)/(g)tan theta sec theta`
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