Question

The trajectory of a projectile in a vertical plane is $y=ax-b{x}^2$, where $a$ and $b$ are constants 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 projection from the horizontal are

• A. $\frac{b^2}{2a},ta{n}^{-1}\left(b\right)$
• B. $\frac{a^2}{b},ta{n}^{-1}\left(2a\right)$
• C. $\frac{a^2}{4b},ta{n}^{-1}\left(a\right)$
• D. $\frac{2a^2}{b},ta{n}^{-1}\left(a\right)$

Answer 1

The correct option is C $\frac{a^2}{4b},ta{n}^{-1}\left(a\right)$