#### Answer

The component of the plane's velocity that is perpendicular to the ground is 15 m/s

#### Work Step by Step

We can find the slope of the hill;
$tan(\theta) = \frac{3}{100}$
$\theta = arctan(\frac{3}{100})$
$\theta = 1.72^{\circ}$
We then find the component of the plane's velocity that is perpendicular to the ground:
$v_{perp} = v~sin(\theta)$
$v_{perp} = (500~m/s)~sin(1.72^{\circ})$
$v_{perp} = 15~m/s$
The component of the plane's velocity that is perpendicular to the ground is 15 m/s.