Answer
The linear speed as seen by an observer on the ground is $~~340~m/s$
Work Step by Step
We can express the angular speed in units of $rad/s$:
$\omega = (2000~rev/min)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s}) = 209.44~rad/s$
We can find the tangential speed:
$v_t = \omega~r$
$v_t = (209.44~rad/s)(1.5~m)$
$v_t = 314.2~m/s$
We can express the plane's speed in units of $m/s$:
$v_p = (480~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 133.3~m/s$
We can find the linear speed as seen by an observer on the ground:
$v = \sqrt{v_t^2+v_p^2}$
$v = \sqrt{(314.2~m/s)^2+(133.3~m/s)^2}$
$v = 340~m/s$
The linear speed as seen by an observer on the ground is $~~340~m/s$
