A pilot heads her plane with a velocity of 275 km/hr north. If there is a strong wind of 130 km/hr blowing east, what is the magnitude of the velocity of the plane in reference to the ground.

X = 130 km/h

Y = 275 km/h

V^2 = 130^2 + 275^2 = 92,525
V = 304.2 km/h.

To find the magnitude of the velocity of the plane in reference to the ground, we can use vector addition.

First, let's represent the velocity of the plane northwards as a vector. Since the plane is heading north, the velocity vector will be directed upwards, and its magnitude will be 275 km/hr.

Next, let's represent the velocity of the wind blowing eastwards as another vector. Since the wind is blowing east, the velocity vector will be directed towards the right, and its magnitude will be 130 km/hr.

To find the combined effect of the plane's velocity and the wind's velocity, we need to add these two vectors. Since these vectors are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resultant vector.

Using the Pythagorean theorem, we have:
magnitude of resultant vector = square root of ((magnitude of the velocity of the plane)^2 + (magnitude of the velocity of the wind)^2)

Substituting the given values:
magnitude of resultant vector = square root of ((275 km/hr)^2 + (130 km/hr)^2)

Calculating the expression:
magnitude of resultant vector = square root of (75625 km^2/hr^2 + 16900 km^2/hr^2)
magnitude of resultant vector = square root of (92525 km^2/hr^2)
magnitude of resultant vector ≈ 304.15 km/hr (rounded to two decimal places)

Therefore, the magnitude of the velocity of the plane in reference to the ground is approximately 304.15 km/hr.