Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and move with velocities $$v1$$ $$=$$ $$3.0$$ $$m/s$$ and $$v2$$ $$=$$ $$4.0$$ $$m/s$$ horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Question

Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and move with velocities $$v1$$ $$=$$ $$3.0$$ $$m/s$$ and $$v2$$ $$=$$  $$4.0$$ $$m/s$$ horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Infinity Learn · Accepted Answer

Let the particle move perpendicular to each other at time t.Here (4i$$⏞$$-gtj$$⏞$$).$$(-3j$$⏞$$-gtj$$⏞$$)$$ $$=$$ $$0$$⇒ $$-12$$ $$+$$ $$g2t2$$ $$=$$ $$0$$⇒ t $$=$$ (12100)  ⇒ t $$=$$ $$35Hence$$ distance,d $$=$$ $$(4t+3t)2+((h-12gt2)-(h-12gt2))2$$  $$=$$ $$7t$$ $$=$$ $$735$$  m

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