Question

A particle is projected from ground with speed 80 m/
s at an angle 30° with horizontal from ground. The
magnitude of average velocity of particle in time
interval t = 2 s to t = 6 s is [Take g = 10 m/s]
(1) 40/2 m/s (2) 40 m/s
(3) Zero
-T41 40/3 m/s​

Answer 1

Answer:

(AV) =total distance÷total time

time=4s (given)

s=Range=u^2sin2x°÷g

80×80×sin30×cos30÷10

6400×1÷2×√3÷2÷10

1600√3÷10

160√3

average velocity=160√3÷4

40√3m/s