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Q.

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.

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a

5 m

b

73 m

c

735m

d

72m

answer is C.

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Detailed Solution

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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