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Question: Answered & Verified by Expert
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
PhysicsMotion In Two DimensionsAP EAMCETAP EAMCET 2014
Options:
  • A $\frac{2 a^2}{b}, \tan ^{-1}(a)$
  • B $\frac{b^2}{2 a}, \tan ^{-1}(b)$
  • C $\frac{a^2}{b}, \tan ^{-1}(2 b)$
  • D $\frac{a^2}{4 b}, \tan ^{-1}(a)$
Solution:
1116 Upvotes Verified Answer
The correct answer is: $\frac{a^2}{4 b}, \tan ^{-1}(a)$
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 $\frac{a^2}{4 b}, \tan ^{-1}(a)$

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