An airplane has an airspeed of 724 kilometers per hour at a bearing of 30 degrees.The wind velocity is 32 kilometers per hour from the west. What are the groundspeed and the direction of the plane?

make a diagram.

from the origin draw a vector with a bearing of 30 degrees and mark its length 724. Call it OA
Draw a horizontal line AB of length 32
the angle OAB = 120 degrees

by cosine law:
OB^2 = 724^2+32^2-2(724)(32)cos120
= 548368
OB= 740.5

now by the Sine Law:
sinO/32 = sin120/740.5
sinO= .0374235
angleO=2.145
so the bearing is 30+2.1 or 31.1 degrees, and the ground speed is 740.5 km/h

check my arithmetic

Well, it seems like this airplane is playing a game of "Where's the Wind?". So, let's calculate the groundspeed and direction for this adventurous plane!

To find the groundspeed, we need to consider the effect of the wind on the plane's airspeed. It's like a strong gust of wind trying to slow down a speeding airplane.

First, let's break down the wind's velocity into its north-south (vertical) and east-west (horizontal) components. Since the wind is coming from the west, it has a velocity of 32 kilometers per hour directly towards the east.

Now, let's use some trigonometry magic to find the horizontal component of the airplane's airspeed. To do that, we'll multiply the airspeed (724 kilometers per hour) by the cosine of the angle (30 degrees):

Horizontal component = 724 km/h * cos(30 degrees)

And since cos(30 degrees) is approximately 0.866, we can calculate:

Horizontal component = 724 km/h * 0.866 ≈ 627.784 kilometers per hour

Now, let's factor in the effect of the wind. Since the wind is pushing the plane to the east, we need to subtract the wind's horizontal component:

Groundspeed = Horizontal component - Wind's horizontal component

Groundspeed = 627.784 km/h - 32 km/h = 595.784 kilometers per hour

So, the groundspeed of the plane is approximately 595.784 kilometers per hour.

Now, let's find the direction of the plane. We'll need to calculate the angle between the groundspeed and the north direction. This angle is also known as the drift angle.

To do that, we'll use the tangent of the angle:

Tangent of drift angle = Wind's vertical component / Wind's horizontal component

Since the wind's vertical component is zero (as it's blowing from west to east), the drift angle is zero. This means the plane's direction is the same as its heading.

Therefore, the direction of the plane is 30 degrees (north of east).

I hope this helps you navigate the windy sky with a little bit of mathematical humor! Enjoy the flight!

To find the groundspeed and direction of the plane, we need to consider the effect of the wind on the plane's motion.

Step 1: Break down the wind velocity into its horizontal and vertical components.
The wind is coming from the west, which means it is blowing towards the east. Since we only need the horizontal component, we do not consider the vertical component. The horizontal component of the wind velocity can be found using trigonometry:
Horizontal component = Wind velocity * cos(Angle between the wind and the plane's heading)
Horizontal component = 32 km/h * cos(180° - 30°)
Horizontal component = 32 km/h * cos(150°)
Horizontal component = 32 km/h * (-0.866)
Horizontal component = -27.712 km/h
(Note: We use a negative sign because the wind is blowing in the opposite direction of the plane's heading.)

Step 2: Find the resultant velocity of the plane relative to the ground.
The resultant velocity is the vector sum of the plane's airspeed and the wind's horizontal component:
Resultant velocity = Airspeed + Horizontal component of wind velocity
Resultant velocity = 724 km/h + (-27.712 km/h)
Resultant velocity = 696.288 km/h

Step 3: Find the direction of the plane.
We can use trigonometry to find the angle of the resultant velocity relative to the north direction. In this case, we will find its bearing relative to the north.
Angle = arccos(Horizontal component / Resultant velocity)
Angle = arccos(-27.712 km/h / 696.288 km/h)
Angle = arccos(-0.039756)
Angle = 91.593°

Since the wind is blowing towards the east, the resultant velocity (groundspeed) of the plane will have a bearing of 30° + 180° = 210° relative to the north.

So, the groundspeed of the plane is approximately 696.288 km/h, and the direction (bearing) of the plane is approximately 210° relative to the north.

To find the groundspeed and direction of the plane, we need to calculate the resultant velocity vector by considering both the airspeed and wind velocity.

1. Start by drawing a diagram to visualize the situation. Draw a horizontal line to represent the direction from where the wind is coming (west) and mark it as "W". Then, draw a line at a 30-degree angle to represent the direction of the plane's airspeed, and mark it as "A" for airspeed.

2. Now, we need to find the components of the airspeed vector. The airspeed can be thought of as having two components: one in the north-south direction and one in the east-west direction. To calculate these components, we use trigonometric functions.

The north-south component (NS) can be found by multiplying the airspeed (724 km/h) by the cosine of the angle (30 degrees):
NS = 724 km/h * cos(30°)

The east-west component (EW) can be found by multiplying the airspeed (724 km/h) by the sine of the angle (30 degrees):
EW = 724 km/h * sin(30°)

3. Now, let's calculate the components of the wind vector. Since the wind is coming from the west, its only component is in the east-west direction:
EW_wind = 32 km/h

4. To find the groundspeed vector, we need to add the components of the airspeed vector and wind vector:
NS_ground = NS + 0 (since the wind does not affect the north-south direction)
EW_ground = EW + EW_wind

5. Finally, we can calculate the magnitude of the groundspeed vector (GS) using the Pythagorean theorem:
GS = √(NS_ground^2 + EW_ground^2)

6. To find the direction of the groundspeed vector, we use trigonometry to calculate the angle between the groundspeed vector and the east-west axis:
Direction = arctan(NS_ground / EW_ground)

Now, let's plug in the values and calculate the groundspeed and direction of the plane:

NS = 724 km/h * cos(30°) = 626.89 km/h
EW = 724 km/h * sin(30°) = 362.00 km/h
EW_wind = 32 km/h

NS_ground = 626.89 km/h
EW_ground = 362.00 km/h + 32 km/h = 394.00 km/h

GS = √(NS_ground^2 + EW_ground^2) = √(626.89^2 + 394.00^2) ≈ 746.22 km/h

Direction = arctan(NS_ground / EW_ground) = arctan(626.89 km/h / 394.00 km/h) ≈ 57.69 degrees

Therefore, the groundspeed of the plane is approximately 746.22 km/h and the direction is approximately 57.69 degrees. This means that the plane is moving at a speed of 746.22 km/h in the direction of approximately 57.69 degrees from the east.

739 kilometers per hour. I used an online electronic flight computer.