In this problem the choice of coordinates system is the most crucial point. Calculations will be easy or difficult depending on the choice of the coordinate system.
Method-I choosing the coordinate system with x-axis in the horizontal direction and y-axis in the vertical direction
For this coordinate system we have
` u_(x) = v, u_(y) = 0 and a_(x) = 0, a_(y) = - g`
At t =0 , x =0 , y =0
` At t=t, x = AB cos theta , y = - AB sin theta`
(a) calculation of AB
For motion along x-axis , AB ` cos theta = vt`
For motion along y-axis, ` - AB sin theta = 0 xx t + 1/2 (-g) t^(2)`
From equations (i) and (ii)
` AB sin theta = g/2 (AB cos theta)/v)^(2)`
` Rightarrow AB = ( 2 v^(2) sin theta)/( gcos^(2) theta)`
` Rightarrow AB = ((2v^(2))/g) tan theta sec theta`
(b) Time taken by the projective to hit the plane
` t = (A B cos theta)/v = ( (2v^(2) tan theta sec theta)/g cos theta)/v`
` Rightarrow t = ( 2 v tan theta)/g `
We can also find the time t by dividing equation (ii)by equaiton (i)
` (AB sin theta)/(AB cos theta) = ( 1/2 gt^(2))/(vt)`
` Rightarrow tan theta = (gt)/(2v)`
` Rightarrow t = ( 2v tan theta)/g`
(c ) Velocity of the projectile just before it hits the plane
Let v be the velocity with which the projectile hits the palne, ` v_(x) and v_(y)` are its components as shown in the figure.
` v_(x) =v` ( horizontal component remains same)
` v_(y) = U_(y) + a_(y) t = 0 + ( -g) t = -g xx ( 2 v tan theta)/g = - 2v tan theta`
Required velocity
` v= sqrt(v_(x)^(2) + v_(y)^(2)) = sqrt(v^(2) + ( - 2 v tan theta)^(2)) = v sqrt(1+ 4 tan^(2) theta)`
The velocity makes and angle ` beta ` with x -axis given by ` beta = tan ^(-1) (|v_(y)|)/(|v_(x)|) = tan^(-1) ( 2 tan theta)`
