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
(4) 40√3 m/s​

Answer 1

Answer:

40√3 m/s

Explanation:

Time of flight = 2usinθ/g

             = 2(80)sin(30)/10

             = 2(8)(1/2)

             = 8 sec

Since the trajectory remains symmetric, particle is at same height t = 2 and t = 6.   There is no vertical displacement.

Horizontal range = velocity x time

   = u cosθ x (given time gap)

   = 80 x cos30 x  (6 - 2)

   = 80 x √3/2 x 4

   = 160√3  m

Average speed = displacement/time

                       = 160√3 / (6 - 2)

                       = 40√3