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
please answer correctly ​

Answer 1

Answer:

Time of flight = 2usintheta/g=

2×80×sin30 /10

=8

As particle during journey at t=2sec and at t=6sec will be at same height and hence vertical displacement is zero.

Horizontal displacement =(ucosθ)×(6−2)

=80× √3/2×4 =160 √3 m

∴ Average velocity = 160 √3\4

=40 √3m/sec

solution