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