Airplane vector when descending

Aug 2016
Plymouth, MA
I have a question from my chapter on vectors:

Find the component form of the vector that represents the velocity of an airplane descending at a speed of 100 mph at and angle of $30^\circ$ below the horizontal.

For the angle of descent I calculated $0^\circ - 30^\circ = -30^\circ = 330^\circ$

The answer at the back of the book calculated the vector using $180^\circ + 30^\circ = 210^\circ$ so my answer was wrong.

But, wouldn't that vector have the airplane going backward? It seems to me the x component of the vector would be positive while the y component would be negative, putting it in Quadrant IV.

Where is my thinking wrong on this?

BTW, my answer was $v=\langle 50\sqrt{3}, -50 \rangle$ while the textbook says it should be $v=\langle -50\sqrt{3}, -50 \rangle$


MHF Helper
Nov 2013
If this is the entire problem it's awful.

You don't seem to be given any information on the x and y directions of motion. Is it just assumed to be along the positive x axis?

Assuming that's the case. You have positive velocity in the x direction and negative velocity in the z direction.

Using the numbers given and your ability to visualize you should be able to see that

$v_x = 100 \cos(30^\circ)=50\sqrt{3}$

$v_z = -100 \sin(30^\circ)=-50$

and thus the velocity vector $v = (50\sqrt{3},~-50)$ which appears to be exactly what you got.

Apparently the plane is travelling along the negative x axis, though why you'd know this I have no idea.
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Aug 2016
Plymouth, MA
I agree Romsek. I should have put quotes around the question. That's exactly how it appears in the textbook.

I'll report it as an error and see what happens. I sure wouldn't want to be on any airplane they were piloting.