Question

The path of a projectile is given by the equation $ y = ax - bx^{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,
$y=a x-b x^{2}\,\,\,...(i)$
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
$\frac{a^{2}}{b}=\frac{\tan ^{2} \theta}{g} \times 2 u^{2} \cos ^{2} \theta$
$=\frac{\sin ^{2} \theta}{g \cos ^{2} \theta} \cdot 2 u^{2} \cos ^{2} \theta$
$=\frac{2 u^{2} \sin ^{2} \theta}{g}=4\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right)$
$H_{\max }=\frac{a^{2}}{4 b}$
and $a=\tan \theta $
$\theta=\tan ^{-1}(a)$
So, required solution is $\frac{a^{2}}{4 b}, \tan ^{-1}(a)$