I have a question regarding two equations.
Imagine an object traveling at 1 m/s and 90 degrees relative to the surface, such that the x-component is in the direction perpendicular to the surface. In the 90 degree case, the y component is 0. Thus the x-component is 1 m/s. If I wanted to have the angle between the object and surface be 89 degrees, the velocity needs to have a y-component. How do I calculate the y-component associated with a degree change of 1 degree?
First I know that:
$$\sqrt{x^2+y^2} = 1 \text{ m/s}$$
and I think:
$$\tan(\theta) = x/y$$
Maybe its degrees or radians I am dealing with, but I would appreciate some guidance in this problem
It's very strange that you would say "I wanted to have the angle between the object and surface be 89 degrees" and then say "Maybe its degrees or radians I am dealing with"! Since you are given the angle in degrees, of course you are dealing with degrees.
Yes, since the distance from the origin to (x, y) is 1, $\sqrt{x^2+ y^2}= 1$ Squaring both sides, $x^2+ y^2= 1$. Yes, $\frac{x}{y}= tan(89)= 57.3$ so that x= 57.3y and then $x^2+ y^2= (57.3y)^2+ y^2= 3284.29y^2= 1$. So $y= \sqrt{\frac{1}{3284.29}}= 0.01745$