Q156 / 160 · TS EAMCET 2014 · Physics

The path of a projectile is given by the equation $y=a x-b x^2$, 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
    $\frac{2 a^2}{b}, \tan ^{-1}(a)$
  • B
    $\frac{b^2}{2 a}, \tan ^{-1}(b)$
  • C
    $\frac{a^2}{b}, \tan ^{-1}(2 b)$
  • Correct answer
    D
    $\frac{a^2}{4 b}, \tan ^{-1}(a)$

Solution

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

Hence, the correct answer is option: 'D'.

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