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 horizontal and vertical distances respectively 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. $\frac{a^2}{b},{\tan }^{-1}\left(2a\right)$
• B. $\frac{2a^2}{b},{\tan }^{-1}\left(a\right)$
• C. $\frac{b^2}{2a},{\tan }^{-1}\left(a\right)$
• D. $\frac{a^2}{4b},{\tan }^{-1}\left(a\right)$

Answer 1

The correct option is D

$\frac{a^2}{4b},{\tan }^{-1}\left(a\right)$

Given,
$y=ax-b{x}^2$

Comparing with

$y=\left(\tan \theta \right)x-\frac{g(1+{\tan }^2\theta )x^2}{2u^2}$

we get:

$\left(i\right)$ $a=\tan \theta$


$\Rightarrow \sin \theta =\frac{a}{\sqrt{1+a^2}}$

Also, $\theta ={\tan }^{-1}a$

$\left(ii\right)$ $b=\frac{g(1+{\tan }^2\theta )}{2u^2}$

Substituting the value, $\tan \theta =a$, we have,

$u^2=\frac{g(1+a^2)}{2b}$

Now, the maximum height of the given projectile,

$h_{\text{max}}=\frac{u^2{\sin }^2\theta }{2g}$

Substituting the values in the above equation, we get

$h_{\text{max}}=\frac{(1+a^2)}{4b}\frac{a^2}{1+a^2}$

$\therefore h_{\text{max}}=\frac{a^2}{4b}$

Hence, option (c) is correct answer.