Question

A particle is projected from ground with speed 80m/s at angle 30° with horizontal from ground.the magnitude of average velocity of particle in time interval t=2s to t=6s is

Answer 1

The time of flight = $\frac{2usin\theta}{g} = \frac{2\times 80\times sin 30}{10} = 8 s.$.
So at 4 seconds the body is at the higest point.
At t=2s and t= 6 second the body is at the same height. So Vertical displacement between 2s and 6 s is zero.( Body went up and returned to the same vertical level.
So vertical average velocity for this period is zero. 
Horizontal velocity during this period is constant which is equal to $u \:cos\theta = 80\times cos 30 = 40\sqrt{3} \:ms^{-1}$.
This is the average velocity between t = 2 s and t = 6 s.

Answer 2

Explanation:

Time of flight = 2u sin theta/g

= 2×80×sin30 / 10

= 8s

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

Average velocity = 160✓3/4 = 40✓3 m/s