Question

Single Correct Answer Type
The path of a projectile is given by the equation $y=ax-b{x}^2$, 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:

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

Answer 1

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

The trajectory equation $y=ax-b{x}^2$

Differentiating w.r.t $x$, $\frac{dy}{dx}=a-2bx$

For maximum height, $\frac{dy}{dx}=0$

$\therefore$ $a-2bx=0$ $⟹x=\frac{a}{2b}$

Thus maximum height $H=y{|}_{x=\frac{a}{2b}}$

$\therefore$ $H=a\times \frac{a}{2b}-b\frac{a^2}{4b^2}=\frac{a^2}{4b}$

Angle of projection is equal to the slope of trajectory at $x=0$

$\therefore$ $tan\theta =$ Slope $=\frac{dy}{dx}{|}_{x=0}$

$\therefore$ $tan\theta =a-2b\left(0\right)=a$

$⟹$ $\theta =ta{n}^{-1}\left(a\right)$