I have trouble with homework question i can't seem to figure out. I have tried have method below to solving. I am not sure am i correct because my answer is wrong.
A jet is heading due east: its nose points towards the east direction, but its trajectory on the ground deviates from the east direction due to a sideways component of the wind. The plane is also climbing at the rate of 100 km/h (height increase per unit time). If the plane's airspeed is 510 km/h and there is a wind blowing 90 km/h to the northwest, what is the ground speed of the plane?
$$c^2=510^2-90+2\cdot \sqrt{510-100}\cdot \cos 45°$$
This question is so old as to be probably useless to the person who asked it, but the author of the earlier answer has asked for an answer that shows why the airspeed vector is not simply the composition of three orthogonal vectors with magnitudes equal to the ground speed, wind speed, and vertical speed, respectively.
Airspeed is the magnitude of velocity vector of the airplane relative to the surrounding air mass. Wind describes the motion of the air mass relative to the ground.
The velocity vector $\mathbb v_a$ of the airplane relative to the air mass therefore has magnitude $510$ with vertical component $100$, northward component $0$, and a positive eastward component $v_e$. As these three components are orthogonal, $ 100^2 + 0^2 + v_e^2 = 510^2 $, that is,
$$ v_e = \sqrt{510^2 - 100^2} \approx 500.09999 $$
(not $\sqrt{510 - 100}$, which appears in the question).
The motion of the aircraft with respect to the ground is the composition of its velocity vector with respect to the air mass and the wind velocity vector $\mathbb w$. The direction of $\mathbb w$ is to the northwest and its magnitude is $90$, so it has a positive northward component $45\sqrt2$ and a negative eastward component $-45\sqrt2$.
The motion of the airplane relative to the Earth, $\mathbb v_a + \mathbb w$, therefore has eastward component $v_e - 45\sqrt2 \approx 436.46038$, northward component $45\sqrt2 \approx 63.63961$, and vertical component $100$. The magnitude of this vector is approximately $452.27$, but the ground speed vector is defined as the projection of that vector onto a horizontal plane and the ground speed $v_g$ is the magnitude of the ground speed vector, so we zero out the vertical component and we are left with $$v_g = \sqrt{(\sqrt{510^2 - 100^2} - 45\sqrt2)^2 + 2(45^2)} \approx 441.075575. $$
Alternatively, we can eliminate the vertical component immediately to arrive at the horizontal component of velocity relative to the air mass. This is a vector of magnitude $\sqrt{510^2 - 100^2}$ east. Add the wind vector $\mathbf w$ to this. The groundspeed, which is the magnitude of the sum of these vectors can then be found by the Law of Cosines with angle $45^\circ$:
$$ v_g^2 = 90^2 + (510^2 - 100^2) - 2 (90)(\sqrt{510^2 - 100^2})\cos 45^\circ. $$
This arrives at the same value of $v_g$ as before.
We can also combine the ground speed, wind speed, and vertical speed somewhat as suggested in the earlier answer, provided that we first convert these into velocity vectors $\mathbf v_g$, $\mathbf w$, and $\dot{\mathbf h}$. The total velocity vector relative to the Earth is $\mathbf v_g + \dot{\mathbf h}$ (remembering that $\mathbf v_g$ is only the horizontal component of the velocity), so $\mathbf v_g + \dot{\mathbf h} = \mathbf v_a + \mathbf w$ and therefore $$ \mathbf v_g + \dot{\mathbf h} - \mathbf w = \mathbf v_a . $$
It is a fact that $\mathbf v_g$ is orthogonal to $\dot{\mathbf h}$ and also a fact that $\dot{\mathbf h}$ is orthogonal to $\mathbf w$, but (as it turns out) $\mathbf v_g$ is not orthogonal to $\mathbf w$. Therefore the magnitude of the sum of these three vectors is not simply the square root of the sum of their squared magnitudes.
Worse still, we don't initially know the direction of $\mathbf v_g$. So it is better to write $$ \mathbf v_g = \mathbf v_a - \dot{\mathbf h} + \mathbf w, $$ because then we know the directions of all three vectors in the sum and can add them correctly (but still accounting for the fact that $\mathbf v_a$ is not orthogonal to $\mathbf w$, so we can't just apply the Pythagorean theorem to the magnitudes of these three vectors). This equation can be resolved by one of the other two methods.