A high-performance jet plane, practicing radar avoidance maneuvers, is in horizontal flight of *h* = 45 m above the level ground. Suddenly, the plane encounters terrain that slopes gently upward at 4.3°, an amount difficult to detect (see Fig. 2-16). How much time does the pilot have to make a correction to avoid flying into the ground? The speed of the plane is1300 km/h.

You will have different numbers than mine for this problem, but the principle of solving it is the same. First, let’s consider the scenario. Is the plane changing it’s speed?

If you answered “No” you win a prize. Since there is no acceleration in the scenario, the actual physics work is very simple. We will be playing with the defn. of a constant velocity:

v=(delta)x/t and we will be solving for t, as per the problem.

The rest of the problem is just geometry and dimensional analysis.

For the geometry:

Consider the slope as a right triangle, the height of which is defined by the height that the plane is flying. The angle opposite that leg is the 4.3 degree slope (from my problem) and we have an unknown hypotenuse, and unknown adjacent leg. The adjacent leg is the key. If the plane flies that far, it will crash into the hill. So we need to find the length of that leg.

Using tangent ooperator:

tan 4.3 = 45/adj

adj = 45/tan 4.3 = 598.5 m

Now we know how far the plane can fly before it runs into the hill.

We also know the speed, but we must convert to m/s. My conversion results in a velocity of 361.1 m/s

Now the physics:

v=(delta)x/t

t = (delta)x/v = 598.5 / 361.1 = **1.66 s**