a helicopter has a speed relative to the air of 76km/h northeast. The speed of the wind relative to the ground is 28km/h south. 2 hours after leaving the ground, what is the displacement of the helicopter relative to the ground?

To find the displacement of the helicopter relative to the ground, we need to consider its velocity relative to the ground.

The helicopter's velocity relative to the ground is the vector sum of its velocity relative to the air and the velocity of the wind relative to the ground. We can calculate this using vector addition.

Given:
- Helicopter's speed relative to the air: 76 km/h (in the northeast direction)
- Wind's speed relative to the ground: 28 km/h (in the south direction)

Step 1: Resolve the helicopter's speed relative to the air into its components.
Since the helicopter is moving in the northeast direction, we can break down its velocity into its eastward and northward components using trigonometry.

Let's assume the eastward component is Vx and the northward component is Vy.
Using the Pythagorean theorem, we can find Vx and Vy:
Vx = speed * cos(angle) = 76 km/h * cos(45°) = 53.74 km/h (approximately)
Vy = speed * sin(angle) = 76 km/h * sin(45°) = 53.74 km/h (approximately)

Step 2: Determine the total velocity of the helicopter relative to the ground.
The eastward component of the wind's velocity is 0 km/h since it is blowing directly south.
Therefore, the total eastward component of the helicopter's velocity relative to the ground is Vx + 0 = 53.74 km/h.

The northward component of the helicopter's velocity relative to the ground is Vy - wind's speed = 53.74 km/h - 28 km/h = 25.74 km/h (approximately) north.

Step 3: Calculate the displacement of the helicopter relative to the ground after 2 hours.
Since displacement is the product of velocity and time, we can multiply the velocity components by the time (2 hours) to find the displacement.

The eastward displacement = (eastward velocity) * (time) = 53.74 km/h * 2 h = 107.48 km eastward.

The northward displacement = (northward velocity) * (time) = 25.74 km/h * 2 h = 51.48 km northward.

Therefore, the displacement of the helicopter relative to the ground after 2 hours is approximately 107.48 km eastward and 51.48 km northward.