- #1

- 19

- 0

I don't quite get, where does the V0x=v * d/(sqrt(d^2+(h-y0)^2) come from?

also, what does (h-y0) equal to? t/v0y?

- Thread starter dramadeur
- Start date

- #1

- 19

- 0

I don't quite get, where does the V0x=v * d/(sqrt(d^2+(h-y0)^2) come from?

also, what does (h-y0) equal to? t/v0y?

- #2

- 32,884

- 11,380

It's the Pythagorean theorem; the velocity ##v## is being split up into its horizontal and vertical components. The ratio of the components will be the same as the ratio of the horizontal and vertical lengths of the triangle in the diagram to the length of the hypotenuse.where does the V0x=v * d/(sqrt(d^2+(h-y0)^2) come from?

- #3

- 32,884

- 11,380

At a particular time ##t##, yes. But that's because ##t## changes, not because ##h - y_0## changes. ##h - y_0## is a constant and is taken straight off the diagram; it's just the difference in two heights (which means it's the vertical side of the triangle I referred to in my previous post).what does (h-y0) equal to? t/v0y?

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