Question

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

Answer: (D)

The given equation,
$y=ax-b{x}^2...(i)$
and we know that equation of trajectory is
$y=(\tan \theta )\times -\frac{1}{2}\frac{g}{u^2{\cos }^2\theta }\cdot x^2...(ii)$
Compare both equations, we get
$a=\tan \theta ,b=\frac{1}{2}\cdot \frac{g}{u^2{\cos }^2\theta }$
The maximum height
$\frac{a^2}{b}=\frac{{\tan }^2\theta }{g}\times 2u^2{\cos }^2\theta$
$=\frac{{\sin }^2\theta }{g{\cos }^2\theta }\cdot 2u^2{\cos }^2\theta$
$=\frac{2u^2{\sin }^2\theta }{g}=4\left(\frac{u^2{\sin }^2\theta }{2g}\right)$
$H_{\max }=\frac{a^2}{4b}$
and $a=\tan \theta$
$\theta ={\tan }^{-1}(a)$
So, required solution is $\frac{a^2}{4b},{\tan }^{-1}(a)$