Two particles are located on a horizontal plane at a distance `60 m`. At `t = 0` both the particles are simultaneously projected at angle `45^@` with velocities `2 ms^-1 and 14 m s^-1`, respectively. Find
(a) Minimum separation between them during motion.
(b) At what time is the separation between them minimum ?
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Explanation

In relative motion, observer consioders himself at rest and observes the motiob of object. Graphically, we can draw the direction of motion of particle `2` w.r.t. particle `1` (Fig. 5.123). Both the particles are moving in gravitation field with same acceleration `g`. Hence, relative acceleration of particle `2` as seen from particle `1` will be zero. It means the relative velocity of the particle `2` w.r.t. particle `1` will be constant and will be equal to initial relative velocity. Graphically, we can draw the situation as shown in (Fig. 5.123).
`AN` is the minimum separation between the particles and `BN` is the relative separation between the particles when the distance between `1 and 2` is shortest. From figure, we can write
`v_(12) cos theta = 14 cos 45^@ + 2 cos 45^@` ...(i)
`v_(12) sin theta = 14 sin 45^@ - 2 sin 45^@` ....(ii)
From (i) and (ii), `v_(12) = 10 sqrt(2) m s^-1`
`cos theta = (4)/(5) and sin theta = (3)/(5), as theta = 37^@`
Hence, minimum separation between the particles
=`AN = AB sin theta = 60 xx (3)/(5) = 36 m`
The time when separation between the particles is minimum,
`t = ( B N)/(|vec v_(12)|) = t = (60 cos 37^@)/(10 sqrt(2)) = (12 sqrt(2))/(5) s`.
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