One second after the projection, a stone moves at an angle of ` 45@` with the horzontal. Two seconds from the start, it is travelling horizontally. Find the angle of projection with the horizontal. (g=10 ms^(-2))`.

Explanation

After one second, let ` v_x. v_y` be the horizontal and vertical coponent velocites of the projectile whose initial velcoity of projection (u) and angle of projection os ` theta` then
` v_x = u cos theta ` and ` v_y =u sin theta -g xx 1 =u sin theta-g`
As the resultant of ` vec _x and vec v_y` makes an angle ` beta (= 45^@) with the horizontal. so
` tan beta =v_y/v_x = ( u sin theta-g)/( u cos theta) =tan 45^@` or ` u sin theta- g = u ` or ` u (sin theta - cos theta ) = g ` ....(i)
After two seconds, the vetical component velocity of projectile becomes zero, since the velocity or projectile is horizontal agter tow seconds . So, ` u sin theta - 2 g = 0 ` or u 2 g/sin theta`
From (i) , (2 g)/( sin theta) (sin theta-cos theta) = g` or ` 2 ( 1- cot theta ) = ` or ` 1-cot theta = 1/2`
or ` cot theta 1- 1/2=1/2 ` or tan theta = 2 ` or ` thata = tan^(-10 (2)`.