An airplane is heading due east with an airspeed of 220km/h when it encounters a wind from the south at 76km/h. What is the direction of the resultant velocity of the airplane
The answer is E19.06°N but I don't know how
Draw a diagram of the forces.
tanθ = 76/220
what is arctan(76/220), this is the angle N of E (remember a wind from the S blows N)
Thanks! I was overthinking the question
Well, we can use a bit of trigonometry to determine the direction of the resultant velocity.
First, let's draw a diagram to visualize the situation. We have the airplane flying east with an airspeed of 220km/h, and we have the wind coming from the south at 76km/h.
Since the airplane is flying directly east, its velocity can be represented as 220km/h due east.
For the wind, since it's coming from the south, its velocity can be represented as 76km/h due north.
Now, to find the resultant velocity, we'll treat the airplane's velocity and the wind's velocity as vectors. We can add the two vectors together to find the resultant vector, which represents the combined effect of the airplane and the wind.
To do this, we'll use a bit of trigonometry. We have a right triangle formed between the airplane's velocity, the wind's velocity, and the resultant velocity. The angle opposite to the resultant velocity is what we're looking for.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
Resultant velocity = sqrt((220^2) + (76^2)) = sqrt(48400 + 5776) ≈ 225.85 km/h
Now, we can use trigonometry to find the angle opposite to the resultant velocity:
tanθ = (opposite/adjacent)
θ = tan^(-1)((76/220))
θ ≈ 19.06°
So, the direction of the resultant velocity is E19.06°N.
To find the direction of the resultant velocity of the airplane, we can use vector addition.
First, let's define the vectors involved:
- The airplane's airspeed vector (A) is heading due east, which we can represent as 220 km/h in the east direction.
- The wind vector (W) is coming from the south, which we can represent as 76 km/h in the north direction.
To find the resultant velocity, we need to add these two vectors together. Since the vectors are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant velocity (R):
R^2 = A^2 + W^2
R^2 = (220 km/h)^2 + (76 km/h)^2
R^2 = 48400 km^2/h^2 + 5776 km^2/h^2
R^2 = 54176 km^2/h^2
R ≈ 232.93 km/h
Now that we know the magnitude of the resultant velocity is approximately 232.93 km/h, we can find its direction.
To find the direction, we can use trigonometric functions. In this case, we'll use the inverse tangent (arctan) function:
tanθ = W/A
θ = arctan(W/A)
θ = arctan(76 km/h / 220 km/h)
θ ≈ 19.06°
Since the wind is coming from the south, the resultant velocity is directed east of north. Therefore, the direction of the resultant velocity of the airplane is E19.06°N.