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The trajectory of a projectile in a vert...

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.

A

`(a^(2))/(4b), tan^(-1)(b)`

B

`(a^(2))/(b), tan^(-1)(2b)`

C

`(a^(2))/(4b), tan^(-1)(a)`

D

`(2a^(2))/(b), tan^(-1)(a)`

Text Solution

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The correct Answer is:
C
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