A particle is projected upwards with velocity 20m//s. Simultaneously another particle is projected with velocity 20(sqrt2) m//s at 45^@. (g = 10m//s^2) (a) What is acceleration of first particle relative to the second? (b) What is initial velocity of first particle relative to the other? (c) What is distance between two particles after 2 s?
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Given :
Velocity of particle 1 (V1) = 20i m/s
Velocity of particle 2 (V2) = 20i + 20j
Angle of projection (∅) = 45°
g = 10 m/s²
To find :
a)acceleration of first particle relative to the second
b)initial velocity of first particle relative to the other
c)distance between two particles after 2 s
Solution :
- a) Acceleration of both the particles is in the downward direction and the magnitude is g.So the relative acceleration is zero
- b)V12 = V1 - V2
= (20j) - (20i+20j)
= -20i
- initial velocity of particle one relative to other is 20m/s in the horizontal direction
- c) V12 = -20i m/s which states that the motion is uniform
Distance= speed × time
= 20 × 2
=40m
The distance between two particles is 40m
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