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A projectile is launched from the origin in the $xy$ plane ( $x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance. $r =\sqrt{ x ^2+ y ^2}$ from the origin is plotted against $x$, the resulting curves show different behaviours for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure below. For $\alpha_1, r ( x )$ keeps increasing with $x$ while for $\alpha_2$, $r(x)$ increases and reaches a maximum, then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at an angle $\alpha_c\left(\alpha_1 < \alpha_c < \alpha_2\right)$. The value of $\alpha_c$ is [ignore where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity]
$\sin ^{-1}\left(\frac{1}{3}\right)$
$\cos ^{-1}\left(\frac{1}{3}\right)$
$\tan ^{-1}\left(\frac{1}{3}\right)$
$\tan ^{-1}(3)$
Solution
(B)
$|\vec{r}|$ will increase if angle between $\frac{d \vec{r}}{d t}$ and $\vec{r}$ is acute.
so $r=\sqrt{ x ^2+ y ^2}$ will always increase if $\vec{r} \cdot \frac{d \vec{r}}{d t} > 0$
$[u \cos \alpha \hat{i}+(u \sin \alpha-g t) \hat{j}]$
$\left[u \cos \alpha t \hat{i}+\left(u \sin \alpha t-\frac{1}{2} g t^2\right) \hat{j}\right] > 0$
$u^2 \cos ^2 \alpha t+u^2 \sin ^2 \alpha t$
$+\frac{1}{2} g^2 t^3-\frac{u \sin \alpha g t^2}{2}-u \sin \alpha g t^2 > 0$
$u^2+\frac{1}{2} g^2 t^2-\frac{3}{2} u \sin \alpha g t > 0$
$\frac{1}{2} g^2 t^2-\frac{3}{2} u \sin \alpha g t+u^2 > 0$
for this quadratic expression to be always positive
$D < 0$
$\frac{9}{4} u^2 \sin ^2 \alpha g^2-2 g^2 u^2 < 0$
$\sin ^2 \alpha < \frac{8}{9} \Rightarrow \sin \alpha < \frac{2 \sqrt{2}}{3}$
$\Rightarrow \cos \alpha > \frac{1}{3}$
