The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

Answer

`a^2/(4b)`

Explanation

Given that `y = ax - bx^2`
Comparing it with equation of a projectile , we get

`y = x tan theta - (gx^2)/(2u cos^2theta) rArr tan theta = a`………(i)
and `g/(2bcostheta) = b`……..(ii)
`u^2 = g/(2bcos^2theta) = (g(a^2+1))/(2b)` ........(iii)
`[.: cos theta = 1/sqrt(a^2+1)]`
Maximum height attained , `H = (u^2sin^2theta)/(2g)` .......(iv)
From Eq. (ii) , `g / (2b) = u^2 cos^2 theta`
`rArr g/(2b) = u^2 - u^2sin^2theta`
`rArr u^2sin^2theta = u^2 - g/(2b)`
`rArr u^2 sin^2 theta = (g(a^2+1))/(2b) - g/(2b) = (ga^2)/(2b)` ...........(v)
From Eqs. (iv) and (v), `H = (ga^2)/(2b xx 2g) = a^2/(4b)` and angle of
projection with horizontal is `theta = tan^(-1) (a)`
Alternatively:
`(dy)/(dx) = a- 2bx = 0` (For maximum height)
`rArr x = a/(2b)`
Substituting the value of `x in y = ax - bx^2 ` to find maximum height, `H = a(a/(2b)) - b (a^2/(4b^2)) = a^2/(2b) - a^2/(2b) = a^2/(4b)` .