- #1
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
- Context: Undergrad
- Thread starter dramadeur
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- physics trajectory
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SUMMARY
The discussion clarifies the derivation of the equation V0x = v * d / (sqrt(d^2 + (h - y0)^2)), which is based on the Pythagorean theorem. The equation represents the horizontal component of velocity in projectile motion, where 'v' is the total velocity, 'd' is the horizontal distance, and '(h - y0)' is the vertical height difference. It is established that '(h - y0)' remains constant and represents the vertical side of the triangle formed in the trajectory analysis.
PREREQUISITES- Understanding of projectile motion principles
- Familiarity with the Pythagorean theorem
- Basic knowledge of trigonometric functions
- Concept of velocity components in physics
- Study the derivation of projectile motion equations
- Learn about the role of initial velocity components in trajectory analysis
- Explore graphical representations of projectile motion
- Investigate the effects of air resistance on projectile trajectories
Students of physics, educators teaching projectile motion, and anyone interested in understanding the mathematical foundations of trajectory calculations.
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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