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4.Aparticleisprojectedfrom ground with velocity 40/2 m/s at 45°. Find(a) velocity and(b) displacement of the particle after 2 s. (g 10 m/s2)

Answer»

Given the Velocity is 40√2

so ux = vcos(45) = 40 m/s

and uy = vsin(45) = 40 m/s

now, in 2 seconds

the displacement in x is = ux*t = 40*2 = 80m

and velocity will be same = 40m/s

the displacement in y will be given by s = 40*(2) -1/2*(10)*(2)² = 80-20 = 60m

and velocity after 2 seconds will be Vy = uy-gt =40-(10*2) = 20m/s

so, the net velocity is √40²+20² = 44.72 m/s

and displacement is √80²+60² = 100m



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