Shadows

I guessed an answer by finding the x and y components of the final velocity vector and using the inverse tangent function to find the complement of that angle, then subtracting from 90 degrees, but I don't know if that's right. I got an answer that looked reasonable though (can't remember any numbers, sorry)

That should be right, as long as you divided the x component by the y component. You could have calculated the angle directly (not its complement) by dividing the y component by the x component.

Why are you unsure about the solution?

@Silktree; Cool, thanks for the help :) I wasn't too sure because we never learned how to do it in class, and because I didn't know if by doing that you assumed the trajectory was a triangle instead of a parabola (since I used trig functions and complements of angles which you usually use on triangles in math class...) But in the end I was running out of time, and I wanted partial credit at least. :P

It seems that your teacher didn't do that good a job teaching this unit, then. Even though the trajectory is indeed a parabola, the fact that we're dealing with instantaneous velocity means that we can look at the motion over a short time interval as if it were linear.

[QUOTE=Silktree;bt258014]It seems that your teacher didn't do that good a job teaching this unit, then. Even though the trajectory is indeed a parabola, the fact that we're dealing with [i]instantaneous[/i] velocity means that we can look at the motion over a short time interval as if it were linear.[/QUOTE]

Oh, I get it! Our class does go kind of fast, so we don't really get enough time to properly understand everything. :P
Thanks so much! :)