Question

A particle has displacement of 12meter toward the east and 5 meter towards the north and then 6 metre vertically upwards find the magnitude of the sum of these displacements

Answer 1

(12^2+5^2)^1/2=13m
(13^2+6^2)^1/2=14.3178m
thus magnitude of sum of displacements is
14.3178m
you can also get the answer by
(12^2+5^2+6^2)^1/2=14.3178m

Answer 2

Hi friend,

Resultant displacement = sqrt(122+52+62)

= sqrt (205)

=14.32 m

Hope it helps!