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(D) The initial velocity components are $u_x = u \cos 	heta = 80 \cos 30^{\circ} = 80 	imes \frac{\sqrt{3}}{2} = 40 \sqrt{3} \,m/s$ and $u_y = u \si

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$40 \sqrt{3}$

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(D) The initial velocity components are $u_x = u \cos 	heta = 80 \cos 30^{\circ} = 80 	imes \frac{\sqrt{3}}{2} = 40 \sqrt{3} \,m/s$ and $u_y = u \sin 	heta = 80 \sin 30^{\circ} = 80 	imes \frac{1}{2} = 40 \,m/s$.The position of the particle at any time $t$ is given by $x(t) = u_x t$ and $y(t) = u_y t - \frac{1}{2} g t^2$.At $t_1 = 2 \,s$:$x_1 = (40 \sqrt{3}) 	imes 2 = 80 \sqrt{3} \,m$$y_1 = 40 	imes 2 - \frac{1}{2} 	imes 10 	imes (2)^2 = 80 - 20 = 60 \,m$At $t_2 = 6 \,s$:$x_2 = (40 \sqrt{3}) 	imes 6 = 240 \sqrt{3} \,m$$y_2 = 40 	imes 6 - \frac{1}{2} 	imes 10 	imes (6)^2 = 240 - 180 = 60 \,m$Since $y_1 = y_2$,the displacement vector is $\Delta \vec{r} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} = (240 \sqrt{3} - 80 \sqrt{3}) \hat{i} + 0 \hat{j} = 160 \sqrt{3} \hat{i} \,m$.The time interval is $\Delta t = 6 - 2 = 4 \,s$.The average velocity is $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{160 \sqrt{3}}{4} \hat{i} = 40 \sqrt{3} \hat{i} \,m/s$.The magnitude of the average velocity is $40 \sqrt{3} \,m/s$.

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

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