" A particle is projected from point A on plane "AB," so that "AB=(2u^(2)tan theta)/(g sqrt(3))" in the figure as shown.If "u" is the velocity of projection,find angle "theta

A particle is projected from point A on plane AB, so that AB=(2u^2tantheta)/(gsqrt3) in the figure as shown. If u is the velocity of projection, find angle theta .

A particle is projected from point A on plane AB, so that AB=(2u^2tantheta)/(gsqrt3) in the figure as shown. If u is the velocity of projection, find angle theta .

A particle is projected at an angle theta with an initial speed u .

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 ?

A particle is projected horizontal 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?

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?

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 ?