A high-performance jet plane, practicing radar avoidance maneuvers, is in horizontal flight of h = 35 m above the level ground. Suddenly, the plane encounters terrain that slopes gently upward at 4.3°, an amount difficult to detect (see Figure 2-22). How much time does the pilot have to make a correction to avoid flying into the ground? The speed of the plane is 750 km/h.



So looking at my triangle. 4.3 degrees for the hill.. line z is the hill and line x is the distance from the plane to the hill. Ok right? so to solve for line x i take the
tan(4.3 degrees) = (opp/adjacent)
so that is
tan(4.3deg)= x/35m

and then to solve for x u multiply the 35m and u get:

x= tan4.3deg * 35m = 2.631meters. and right here i know i am wrong cause i know the answer for x is 465.4 meters..

ah ha solved it... incase anyone is wondering aswell here is a reply from a freind on what i did wrong..

"
you drew the picture wrong.
4.3 degrees slope means 4.3 degrees from horizontal in your picture.
That corner angle should be 85.7 degrees, not 4.3
"

Angle is 90 - 4.3 = 85.7 degrees.

x/35 = tan (85.7)

x = (35). tan (85.7)

x= 465.4 metres

Now speed = 750 km/h

= 208.33 m/s

Time taken by plane

= 2.2 seconds.