Two particles are projected simultaneously from two points O and O' such that $$10$$ m is the horizontal and $$5$$ m is the vertical distance between them as shown in the figure. They are projected at the same inclination $$60o$$ to the horizontal with the same velocity $$10$$ $$ms-1$$. Find the time after which their separation becomes minimum.

Question

Infinity Learn · Accepted Answer

Let us find relative velocity of O. w.r.t. O'.

$\begin{array}{l} {\vec{v}}_{olo} = {\vec{v}}_{o} - \vec{v_{o}} = (10 \cos 60^{\circ} \hat{i} + 10 \sin 60^{\circ} \hat{j}) \\ - [10 \cos 60^{\circ} (- \hat{i}) + 10 \sin 60^{\circ} \hat{j}] = 10 \hat{i} \end{array}$

This is along horizontal direction.
If we assume O' to be at rest, then O will move along OA w.r.t. O' and minimum separation will be when particle O is at A.

$For this, relative displacement travelled = O A = 10 m$

$Time taken = \frac{10}{v_{O / O^{'}}} = \frac{10}{10} = 1 s$

Projectile motion

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