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dramadeur
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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?
Dart Trajectory: V0x Calculation Explained
- Thread starter dramadeur
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The equation V0x = v * d / (sqrt(d^2 + (h - y0)^2)) derives from the Pythagorean theorem, where the velocity is divided into horizontal and vertical components. The ratio of these components corresponds to the triangle formed by the horizontal distance and the height difference. The term (h - y0) represents a constant vertical distance between two heights in the diagram. While the time variable t changes, (h - y0) remains fixed. Understanding these relationships is crucial for accurately calculating projectile motion.
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
PeterDonis
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dramadeur said:
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.
- #3
PeterDonis
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dramadeur said:
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).
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