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:

Options

A. `(b^(2))/(2a),tan^(-1)(b)`
B. `(a^(2))/(b),tan^(-1)(2a)`
C. `(a^(2))/(4b),tan^(-1)(a)`
D. `(2a^(2))/(b),tan^(-1)(a)`

Answer

3

Explanation

`y=ax-bx^(2)`
for y to be maximum (H)
`(dy)/(dx)=0`
`a-2bx=0`
`x=(a)/(2b)`
`y_("max")=a((a)/(2b))-v((a)/(2b))^(2)=(a^(2))/(4b)`
`((dy)/(dx))_(x=0)=a=tantheta_(0)`
where `theta_(0)` is the angle of projectile
`theta_(0)=tan^(-1)(a)`