Question

The trajectory of a projectile in a vertical plane is, $y=\alpha x-\beta x^2$, where $\alpha ,\beta$ are constants, $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection and the maximum height attained are respectively given by :

• A. ${\tan }^{-1}\alpha ,\frac{{\alpha }^2}{4\beta }$
• B. ${\tan }^{-1}\beta ,\frac{{\alpha }^2}{2\beta }$
• C. ${\tan }^{-1}\alpha ,\frac{{\alpha }^2}{2\beta }$
• D. ${\tan }^{-1}\beta ,\frac{{\alpha }^2}{4\beta }$

Answer 1

The correct option is A ${\tan }^{-1}\alpha ,\frac{{\alpha }^2}{4\beta }$

Given:
$y=\alpha x-\beta x^2$

Also, $y=x\tan \theta -\frac{g}{2u^2{\cos }^2\theta }{x}^2$

On comparing, we get,

$\tan \theta =\alpha ...\left(1\right)$, and

$\frac{g}{2u^2{\cos }^2\theta }=\beta \Rightarrow \frac{u^2}{g}=\frac{1}{2\beta {\cos }^2\theta }...\left(2\right)$

So, angle of projection, $\theta ={\tan }^{-1}\alpha$
$[$From $\left(1\right)]$

Now, maximum height attained,
$H_{\text{max}}=\frac{u^2{\sin }^2\theta }{2g}$

$\Rightarrow H_{\text{max}}=\frac{{\sin }^2\theta }{4\beta {\cos }^2\theta }$
$[$From $\left(2\right)]$

$\Rightarrow H_{\text{max}}=\frac{{\tan }^2\theta }{4\beta }=\frac{{\alpha }^2}{4\beta }$
$[$From $\left(1\right)]$