What Calculations Determine the Volleyball's Trajectory in Projectile Motion?

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Homework Help Overview

The discussion revolves around the calculations needed to determine the trajectory of a volleyball in projectile motion, specifically focusing on the initial velocity and launch angle required for the ball to clear the net and land at a designated point on the court. The problem includes parameters such as the height from which the ball is served and the distances involved on a volleyball court.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the launch angle, initial velocity, and the trajectory of the ball. Some question the assumptions about the height of the net and the dimensions of the volleyball court. Others discuss the implications of neglecting air resistance and how that affects the trajectory calculations.

Discussion Status

The discussion is ongoing, with various participants offering different methods and interpretations of the problem. Some suggest using known points on a parabola to derive equations, while others emphasize the need for clarity on the highest point of the ball's trajectory. There is recognition of the need to establish the correct assumptions and parameters before proceeding with calculations.

Contextual Notes

Participants note the height from which the ball is thrown (0.9 m) and the assumption that air resistance can be neglected. There is also mention of the need to clarify the dimensions of the volleyball court and the specific points involved in the problem setup.

  • #31
verty said:
I believe Haruspex's method does not require one to differentiate at all which I think is much nicer.
That's very true, I still like more your method for that exact reason, that it involves derivatives!
 
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  • #32
Delta² said:
Not sure what you mean here, but I agree if you correctly treat the system , you can solve one quadratic equation and some linear equations and get the angle and the velocities.
You get two linear equations in which the unknowns are vy/vx and 1/vx2, vx and vy being the initial velocity components.
 
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  • #33
verty said:
Ok, I can see Haruspex's method is the one to use (for sure).
It is no different from your own method in post #21. For some reason I had not seen that when I wrote post #23. There can be quite a delay.
I get two equations instead of three merely by taking coordinates relative to the launch point.
 
  • #34
haruspex said:
It is no different from your own method in post #21. For some reason I had not seen that when I wrote post #23. There can be quite a delay.
I get two equations instead of three merely by taking coordinates relative to the launch point.

I meant working with the usual time formula. Solving those quadratics was not so difficult in this case.
 
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