We will use the trigonometric ratios to know the helicopter's deceleration speed. Later applying the concept of speed as a vector component of the value found we will find the vertical speed.
The 2% grade indicates that for every 100 meters traveled in the x direction, there is an ascent / descent of 2 meters. Therefore we will have the relationship
[tex]\rightarrow \frac{y}{x} = \frac{2}{100}[/tex]
Now you would have the value o[tex]v_y = 6.78m/s[/tex]f the angle tangent would be
[tex]tan \theta = \frac{y}{x}[/tex]
[tex]tan \theta = \frac{2}{100}{[/tex]
[tex]tan\theta = 0.02[/tex]
From this relationship we could conclude that the vertical speed would be
[tex]v_y = v*tan\theta[/tex]
[tex]v_y = 339*0.02[/tex]
[tex]v_y = 6.78m/s[/tex]
Therefore the component of the helicopter's velocity perpendicular to the sloping surface of the hill is 6.78m/s
