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Question

# A particle is projected horizontally with a speed u from the top of a plane inclined at an angle θ0 with the horizontal. How far from the point of projection will the particle strike the plane?

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Solution

## Take X,Y axes as shown in fugure. Suppose that the particle strikes the plane at a point P with coordinates (x,y). Consider the motion between A and P.Now,Motion in x direction =uAcceleration a=0And x=utNowMotion in y direction Initial velocity u=0Acceleration a=gSo,y=ut+12gt2=12gt2y=!2gt2Now eliminate t from x and y respectively, we gety=xtanθ..........(1)Thus,xtanθ=gt22u2 On solving for x get two values i.e.x=0 and x=2u2tanθgTherefore the point is corresponding to x=2u2tanθgSo put this value of x in equation (1) we get,y=xtanθ=2u2tan2θgThen,The distance is OP=√x2+y2 =2u2tanθ√1+tanθg =2u2tanθsecθg

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