A particles is projected from the foot of an inclined plane having inclination `45^(@)` , with the velocity u at angle `theta( gt 45^(@))` with the horizontal in a vertical plane containing the line of greatest slope through the point of projection. Find the value of ` tantheta` if the particle stikes the plane.
(i) Horizontally
(ii) Normally

Explanation

Let A be the point where the particel strikes the plane after time t.
and OB = AB =h
` Rightarrow OA = hsqrt2`
(i) As the paticle strikes the plane horinzontally, vetical component of its velocity at A will be zero.
For horizontal motion.
`OB = ( u cos theta)t Rightarrow h = u cos theta t`
For vertical motion from equation v = u + at
` 0 = u sintheta - gt `
` or usin theta = "gt" `
again, BP ` = ( u sin theta) t - 1/2 "gt"^(2)`
`h = ( U sin theta) - 1/2 "gt"^(2) `
From equations (i),(ii) and (iii)
` u cos thetat = u sin theta t - 1/2 ( u sin tehta) t`
` Rightarrow tan theta = 2`
(ii) As the particle hits the plane normally, component of its velocity parallel to the plane at this instant will be zero and displancement perpendicular to the plane will be zero.
for the motion along the plane,.
From v = u +at
` 0 = u cos ( theta - 45^(@)) - g sin45^(@)t`
` Rightarrw u cos ( theta - 45^(@)) = g/sqrt2 t`
Again , for the motion perpendicular to the plane from s` = ut + 1/2 at^(2`
` 0 = u sin ( theta - 45^(@) ) t - 1/2 g cos 45^(@) t^(2)`
` Rightarrow t= ( 2sqrt2 u)/g sin ( theta - 45^(@))`


From equation (i) & (ii)
` u cos ( theta - 45^(@)) = g/sqrt2 [ (2sqrt2u)/g sin ( theta - 45^(@))]`
` Rightarrow u cos (theta - 45^(@)) = 2 usin ( theta - 45^(@))`
` Rightarrow tan ( theta -45^(@)) = 1/2`
` Rightarrow ( tan theta -1)/(1 + tan theta) = 1/2 `
` Rightarrow tan theta = 3`