1. ## Velocity Vector Problem

There is a plane that just took off and is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a ate of 300 m/min. Resolve its velocity vector into components.

First, I converted 200 m/min to 18 km/hr so all values have the same units. Then I said the velocity of the plane would =

(-) 200 cos135degrees i + 200 sin135degrees j + 18 k

= - 141.42 i + 141.42 j + 18 k

However, the correct answer is - 140.8 i + 140.8 j + 18 k

Can someone please explain where I made my mistake? Thanks very much!

2. There is a plane that just took off and is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity vector into components.
300 m/min = 18 km/hr

angle of climb relative to the x-y plane ... $\displaystyle \theta = \arcsin\left(\frac{18}{200}\right) \approx 5.16^{\circ}$

speed along the x-y plane is $\displaystyle 200\cos{\theta}$

x component = $\displaystyle -(200\cos{\theta})\cos(45^{\circ}) = -140.8 \, km/hr$

y component = $\displaystyle (200\cos{\theta})\sin(45^{\circ}) = 140.8 \, km/hr$

3. Hello, Frostking!

There is a plane that just took off and is climbing northwest through still air
at an airspeed of 200 km/hr, and rising at a rate of 300 m/min.
Resolve its velocity vector into components.

The answer is: .$\displaystyle - 140.8 i + 140.8 j + 18 k$

Look at the side view of the plane climbing.
Code:
    B *
|   *
|       *   200
18 |           *
|               *
|                   *
C * - - - - - - - - - - - * A
x

The hypotenuse $\displaystyle AB = 200,\;BC = 18.$

Hence: .$\displaystyle x\:=\:\sqrt{200^2-18^2} \:=\:\sqrt{39,\!676} \quad\Rightarrow\quad x \:\approx\: 199.19$

Now look down at the ground.
Code:
                |
C * - - - - + E
: *       |
:   * x   |
:     *45°|
:   45° * |
- - + - - - - * - - -
D         |A
|

We see that: .$\displaystyle AD \:=\:AE \:=\:\frac{x}{\sqrt{2}} \:\approx\:140.8$