The airplane can take off when its airspeed (speed of the air flowing over the wing) is equal to 65 knots. What is the length of runway required for the plane to take off if there is a 25 knots head wind? The runway at the Tallahassee Regional Airport has a length of 8000 ft.

If the airplane faces a constant headwind of 25 knots, the airspeed starts at 25. So a minimum ground speed of 65-25=40 knots will be required.

Without a knowledge of the charactistics of the airplane, it would not be possible to calculate the length of the runway required. It depends on the maximum acceleration of the equipment, the loading, etc.

To determine the length of runway required for the plane to take off, we need to calculate the groundspeed and then apply it to the formula for determining the runway length.

To find the groundspeed, we subtract the headwind (25 knots) from the airspeed at takeoff (65 knots):

Groundspeed = Airspeed - Headwind
Groundspeed = 65 knots - 25 knots
Groundspeed = 40 knots

Now that we know the groundspeed, we can use the formula for determining runway length:

Runway Length = Groundspeed * Time

To find the time, we need to convert the length of the runway from feet to the same units as groundspeed (knots), which requires converting feet to nautical miles (NM).

1 NM is equal to 6076.12 feet.

Therefore, the runway length in nautical miles is:

Runway Length (NM) = 8000 ft / 6076.12 ft/NM
Runway Length (NM) = 1.314 NM (rounded to three decimal places)

Now, we can calculate the required runway length:

Runway Length = Groundspeed * Time
8000 ft = 40 knots * Time

To solve for Time, we divide both sides of the equation by 40 knots:

Time = 8000 ft / 40 knots
Time = 200 feet

Finally, we have to convert Time from feet back to nautical miles:

Time (NM) = 200 ft / 6076.12 ft/NM
Time (NM) = 0.0329 NM (rounded to four decimal places)

Therefore, the length of runway required for the plane to take off, considering a 25-knot headwind, is approximately 0.0329 nautical miles or 200 feet.