Question

A jet plane is flying horizontally with a speed of 500 m/s over a hill that slopes upward with a 3% grade (i.e., the "rise" is 3% of the "run"). What is the component of the plane's velocity perpendicular to the ground? (Assume the +x-direction is up along the slope and the +y-direction is perpendicular to the slope and out of the ground.)

Answer 1

Let v=ground velocity of plane = 500 m/s

Since slope is rise/run, so the angle of the "ground" relative to the horizontal
θ=tan<-1>(3/100)

Component of velocity vector perpendicular to the ground
= v*sin(θ)
= ?

Answer 2

That didn't work, I already tried that. I only have 1 more try left so I need to get the right answer this time haha

Answer 3

tan theta = 3/100 so sin theta = .0299865
-500 (.0299865) = -14.99
they defined y as up so it is negative

Answer 4

Thank you, that worked!!

Answer 5

You are welcome.

Answer 6

Good catch, didn't see the details!

Answer 7

Why is it negative?

Answer 8

"Assume the +x-direction is up along the slope and the +y-direction is perpendicular to the slope and out of the ground."

Since the perpendicular component of the velocity "digs" into the ground, so by definition of the y-axis, it is negative.