Question

The trajectory of a projectile in a vertical plane is given by $y=ax-b{x}^2,$ 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. $\frac{b^2}{2a},{\tan }^{-1}\left(b\right)$
• B. $\frac{a^2}{b},{\tan }^{-1}\left(2a\right)$
• C. $\frac{a^2}{4b},{\tan }^{-1}\left(a\right)$
• D. $\frac{2a^2}{b},{\tan }^{-1}\left(a\right)$

Answer 1

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

Trajectory of projectile in a vertical plane is given as
$y=ax-b{x}^{2}y=ax(1-\frac{x}{\left(\frac{a}{b}\right)})....\left(i\right)$
Compare $\left(i\right)$ with $y=x\tan \theta [1-\frac{x}{R}]$
$\Rightarrow \tan \theta =a$ and $R=\frac{a}{b}$
$\therefore \theta ={\tan }^{-1}\left(a\right)$

Now, For maximum height $H,$ we have
maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$ and Range $R=\frac{u^2\sin 2\theta }{g}$
So, we have the ratio $\frac{H}{R}=\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{g}}$
$\Rightarrow R\tan \theta =4H$
$\Rightarrow \frac{a}{b}\times a=4H\Rightarrow H=\frac{a^2}{4b}$