A particle is projected from ground with speed $80m/s$ at an angle ${30}^o$ with horizontal from ground.The magnitude of average velocity of the particle in time interval $t=2s$ to $t=6s$ is [Take $g=10m/s^2$]

The correct option is D $40\sqrt{3}m/s$
Time of flight = $\frac{2u\sin \theta }{g}$

=$\frac{2\times 80\times \sin {30}^o}{10}=8$

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

Horizontal displacement =$\left(u\cos \theta \right)\times (6-2)$

=$80\times \frac{\sqrt{3}}{2}\times 4$

=$160\sqrt{3}m$

$\therefore$ Average velocity = $\frac{160\sqrt{3}}{4}=40\sqrt{3}m/sec$