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(B) The distance $r$ from the origin is given by $r^2 = x^2 + y^2$. For $r$ to increase monotonically with $x$,we require $\frac{dr}{dx} > 0$,which is

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

Vedclass · Answer

(B) The distance $r$ from the origin is given by $r^2 = x^2 + y^2$. For $r$ to increase monotonically with $x$,we require $\frac{dr}{dx} > 0$,which is equivalent to $\frac{d(r^2)}{dt} > 0$ since $x$ increases with time $t$.Given $x = v_0 \cos \alpha \cdot t$ and $y = v_0 \sin \alpha \cdot t - \frac{1}{2}gt^2$,we have:$r^2 = (v_0 \cos \alpha \cdot t)^2 + (v_0 \sin \alpha \cdot t - \frac{1}{2}gt^2)^2$$r^2 = v_0^2 t^2 \cos^2 \alpha + v_0^2 t^2 \sin^2 \alpha - v_0 \sin \alpha \cdot g t^3 + \frac{1}{4}g^2 t^4$$r^2 = v_0^2 t^2 - v_0 \sin \alpha \cdot g t^3 + \frac{1}{4}g^2 t^4$Differentiating with respect to $t$:$\frac{d(r^2)}{dt} = 2 v_0^2 t - 3 v_0 \sin \alpha \cdot g t^2 + g^2 t^3$For $r$ to increase,$\frac{d(r^2)}{dt} > 0$ for all $t > 0$. Dividing by $t$:$g^2 t^2 - 3 v_0 \sin \alpha \cdot g t + 2 v_0^2 > 0$This quadratic in $t$ is always positive if its discriminant $D $D = (-3 v_0 \sin \alpha \cdot g)^2 - 4(g^2)(2 v_0^2) $9 v_0^2 g^2 \sin^2 \alpha - 8 v_0^2 g^2 $\sin^2 \alpha Since $\cos^2 \alpha = 1 - \sin^2 \alpha$,we have $\cos^2 \alpha > 1 - \frac{8}{9} = \frac{1}{9}$.Thus,$\cos \alpha > \frac{1}{3}$. The critical angle is $\alpha_c = \cos^{-1}\left(\frac{1}{3}\right)$.

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