An airplane has a velocity of 454 km/h relative to the moving air. At the same time, a wind blows northward with a speed of 99 km/h relative to earth. The airplane is moving in an easterly direction relative to earth. Find the speed of the airplane to relative to earth.

So, i used pythagorean theorem so square root of 454 squared plus the square root of 99 squared.. and i didn't get the right answer. Can someone explain this problem to me please?

speed=sqrt(454^2+99^2)

123

Certainly! This problem involves vector addition, since the airplane's velocity and the wind's velocity have different directions.

To find the speed of the airplane relative to the earth, we need to find the resultant velocity by adding the velocities of the airplane and the wind.

The velocity of the airplane in the easterly direction is 454 km/h, and the velocity of the wind in the northerly direction is 99 km/h. To find the resultant velocity, we can use vector addition.

Using the Pythagorean theorem, which applies to right triangles, to find the magnitude (or speed) of the resultant velocity, we can use the formula:

Resultant velocity = √(velocity of the airplane)^2 + (velocity of the wind)^2

Plugging in the given values:

Resultant velocity = √(454 km/h)^2 + (99 km/h)^2

Resultant velocity = √(206116 km^2/h^2 + 9801 km^2/h^2)

Resultant velocity = √(216917 km^2/h^2)

Resultant velocity ≈ 466.21 km/h

Therefore, the speed of the airplane relative to the earth is approximately 466.21 km/h.