The path of a projectile is given by the equation `y = ax – bx^(2)`, where a and b are constants and x and y are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively.

Options

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

Answer

(d)

Explanation

`y = ax - bx^(2)`…….(1)
Equation of trajectory, `y =(tan theta )x - 1/2 g/(U^(2)cos^(2)theta)x^(2)` ……….(2)
On compairing equation (1) and (2),
`rArr a = tantheta,b = 1/(2) g/(U^(2)cos^(2)theta) `
Angle of projection, `theta = tan^(-1)a`
The maximum height,
`a^(2)/b=tan^(2)theta/(g)xx2U^(2)cos^(2)theta = 4[U^(2)sin^(2)theta/(2g)]`
`:. H_(max) = a^(2)/(4b)`