A body projected from a point `O` at an angle `theta`, just crosses a wall `y` `m` high at a distance. `x` `m` from the point of projection and strikes the ground at `Q` beyond the wall as shown, then find height of the wall

Explanation

We know that the equation of the trajectory is `y=x tan theta-(gx^(2))/(2u^(2)cos^(2) theta)`can be written as
`y=x tan theta-((gx^(2))/(2u^(2)cos^(2) theta))sintheta/sin theta`
`y=x tan theta-(gx^(2)tan theta)/(u^(2)(2sin theta cos theta)) rArr y=x tan theta-(x^(2)tan theta)/((u^(2)sin2 theta)/(g))`
`rArr =x tan theta [1-(x/R)][ :' R=(u^(2)sin2theta)/(g)]`