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(D) Let the position of the first particle projected from $O$ be $\vec{r}_1(t) = (0, vt - \frac{1}{2}gt^2)$.Let the position of the second particle pr

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$h/2v$

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(D) Let the position of the first particle projected from $O$ be $\vec{r}_1(t) = (0, vt - \frac{1}{2}gt^2)$.Let the position of the second particle projected from $P(0, h)$ be $\vec{r}_2(t) = (vt \cos 	heta, h + vt \sin 	heta - \frac{1}{2}gt^2)$.The relative position vector $\vec{r}_{rel} = \vec{r}_2 - \vec{r}_1 = (vt \cos 	heta, h + vt \sin 	heta - vt) = (vt \cos 	heta, h - vt(1 - \sin 	heta))$.The square of the distance $D^2 = |\vec{r}_{rel}|^2 = (vt \cos 	heta)^2 + (h - vt(1 - \sin 	heta))^2$.To minimize $D^2$,we differentiate with respect to $t$ and set to zero:$\frac{d(D^2)}{dt} = 2(vt \cos 	heta)(v \cos 	heta) + 2(h - vt(1 - \sin 	heta))(-v(1 - \sin 	heta)) = 0$.$v^2 t \cos^2 	heta - hv(1 - \sin 	heta) + v^2 t(1 - \sin 	heta)^2 = 0$.$v^2 t [\cos^2 	heta + (1 - \sin 	heta)^2] = hv(1 - \sin 	heta)$.$v^2 t [\cos^2 	heta + 1 + \sin^2 	heta - 2 \sin 	heta] = hv(1 - \sin 	heta)$.$v^2 t [2 - 2 \sin 	heta] = hv(1 - \sin 	heta)$.$2v^2 t (1 - \sin 	heta) = hv(1 - \sin 	heta)$.$t = \frac{h}{2v}$.

$A$ particle is projected vertically upwards from $O$ with velocity $v$ and a second particle is projected at the same instant from $P$ (at a height $h$ above $O$) with velocity $v$ at an angle of projection $\theta$. The time when the distance between them is minimum is

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