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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
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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 $\frac{a^2}{4 b}, \tan ^{-1}(a)$
$$
y=a x-b x^2
$$
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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