Question

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

• A. $40\sqrt{2}m/s$
• B. $40m/s$
• C. $40\sqrt{3}m/s$
• D. Zero

Answer 1

The correct option is C $40\sqrt{3}m/s$

$u=80,\theta ={30}^o$

$v_x=u_x=ucos\theta$

$v_x=40\sqrt{3}$

$v_y=u_y-gt$

$v_y=40-10t$

From $t=2\to t=6$ finding average value of $v_y$

at $t=2$ , $v_y=20$

at $t=6$ , $v_y=-20$

so average value of $v_y$ From $t=2\to t=6$ $=\frac{20-20}{2}=0$

From $t=2\to t=6$ average value of $v_x$$=40\sqrt{3}$

Therefore averrage velocity$=40\sqrt{3}$