Q. The trajectory of a projectile in vertical plane in $y = ax - bx^{2},$ 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 projection from the horizontal are
A) $\frac{b^{2}}{4b},\tan^{- 1}{(b)}$
B) $\frac{a^{2}}{b},\tan^{- 1}{(2b)}$
C) $\frac{a^{2}}{4b},\tan^{- 1}{(a)}$
D) $\frac{{2a}^{2}}{b},\tan^{- 1}{(a)}$
Explanation
(c) $y = ax - bx^{2}$
For height or $y$ to be maximum
$\frac{dy}{dx} = 0$ or $a - 2bx = 0$ or $x = \frac{a}{2b}$
$\therefore\ \ y_{\max\ } = a\left( \frac{a}{2b} \right) - b\left( \frac{a}{2b} \right)^{2} = \frac{a^{2}}{4b}$
and $\left( \frac{dy}{dx} \right)_{x = 0} = a = \tan\theta$
where $\theta =$ angle of projection
$\therefore\ \theta\tan^{- 1}{(a)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.