A projectile is launched from the origin in t | vedclass

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Question

(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

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{1}{3}\right)$

Vedclass · Answer

(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 $D < 0$ $\frac{9}{4} u^2 \sin ^2 \alpha g^2-2 g^2 u^2 < 0$ $\sin ^2 \alpha < \frac{8}{9} \Rightarrow \sin \alpha < \frac{2 \sqrt{2}}{3}$ $\Rightarrow \cos \alpha > \frac{1}{3}$

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