a plane flies due north at 225km/h. a wind blows it due east at 55km/h. what is the magnitude and direction of the plane's resultant velocity?

okay, one plane is flying north, straight up and the wind blows the plane east [right]. This is almost similar to pythagorean theorem. My guess is to use x^2 + y^2 = z^2. But im pretty sure there's a formula for this in the first or two chapters of the physics book. Look there at use that formula. But in this case you have to find x which is opposite to the wind and plane.

it is pretty simple. to find the magnitude use the pythagorean theorem to find the resultant and then to the tangent inverse to find your angle

231.6km/h

To find the magnitude and direction of the plane's resultant velocity, we can use vector addition.

Let's break down the velocities into their components. The plane's velocity due north can be considered as a vector with only a north component, while the wind's velocity due east can be considered as a vector with only an east component.

The magnitude of the plane's velocity due north is 225 km/h, and the magnitude of the wind's velocity due east is 55 km/h. Since these velocities are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant velocity.

Magnitude of resultant velocity:
Resultant Velocity = sqrt((225 km/h)^2 + (55 km/h)^2)
= sqrt(50625 km^2/h^2 + 3025 km^2/h^2)
= sqrt(53650 km^2/h^2)
≈ 231.77 km/h (rounded to two decimal places)

Now, to find the direction of the resultant velocity, we can use trigonometry. The direction can be determined by finding the angle between the north direction and the resultant vector using the inverse tangent function.

Direction of resultant velocity:
Direction = tan^(-1)((55 km/h) / (225 km/h))

Using a calculator, we can find that the angle is approximately 13.68 degrees (rounded to two decimal places).

Therefore, the magnitude of the plane's resultant velocity is approximately 231.77 km/h, and the direction is approximately 13.68 degrees north of east.