What direction is the pilot flying and what is the speed relative to the ground?

Question

A pilot is flying at an air speed of 241 mph in a wind blowing 20.4 mph from the east. In what direction must the pilot head in order to fly due north?(Round to the nearest tenth degree with a direction E or W from north) What is the pilot's speed relative to the ground?

I guess it has to do with reslutants??

Thanks for any guidance and answers.

Answer 1

Lots of unnecessary complications here.

Airspeed is the hypotenuse. 20.4 is the side opposite the angle we are looking for.

The heading, east of north, must be at angle sin^-1(20.4/241).

Speed relative to the ground is the third side of the triangle, and is

sqrt(241^2 - 20.4^2)

Zidane has got the angles all wrong. The wind is from the east. We want to fly due north. That's where the right angle is. 241 is the hypotenuse of the triangle. The wind serves to make the ground speed less than 241, not greater.

Answer 2

In order to do this question we have to believe that the pilot was flying 241 mph north afterwhich the wind hit at 20.4 mph.You have to use the pythogarian formula where x is the unknown resultant value and the y angle of the resultant.

x=square root(241^2+20.4^2)=241.9m/s is the resultant speed


tan(y)=opp/add

y=tan^-1(inverse)*opp/add=
tan^-1(inverse)*241/20.4=1.49 degrees

Now the second question is solved and the pilots resultant velocity and direction is 242m/s 1.49 degrees N of E.

Now the first question can be easily solved, what better way to counter 20.4 east then to fly its counter measure with the plane heading more west with a resultant value between 20.4 west and 241 north. Threfore to continue to fly north the pilot should head
241.9m/2 1.49 degrees N of W which acts to counter 20.4m/a east wind and the plane will continue to move north when the wind hits.

Answer 3

Vectors!

You want to fly into the wind so that your velocity component in the winds direction cancels out the wind. Your airspeed is your hypothenuse and the side opposite you heading angle is the wind speed. This gives you a sine relationship for your angle.

To help understand. Consider what would happen if you didn't head into the wind.

The speed along the ground is the cosine of the heading angle.

Draw a couple of sketchs, especially assuming that the pilot didn't know there way any wind. See how the plane would go?

Answer 4

Let pilot flies at a speed x in a direction making angleA with the direction of wind, then 241 is the resultant of x &20.4
And angle between x &20.4 is A
And 241 makes an angle 90degree with 20.4
Then
Tan90=(x*sinA)/(20.4+x*cosA)
This means (20.4+x*cosA)=0……(1)
As tan90 is infinity
Now using formula for resultant
(241)^2=x^2+(20.4)^2+2*20.4*x*cosA…..(2)
use value of x*cosA from (1) in(2)
and solve for x andA

Answer 5

Assuming the pilot is correcting for wind shear, he or she is flying at a hundred seventy five kph due south. i do no longer think of you are able to fly 175kph relative to the air...? in any different case, while you're assuming that the airplane is blown off objective, the difficulty will become one you are able to resolve with the Pythagorean Theorem, i could think of. Plug in the two values to the legs of the triangle, and resolve.