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