B Calculating the perfect tennis shot using vectors

[themethetion]

TL;DR Summary

I have developed a function that represents the perfect shot in terms of x & y where x is the court's length and y is the height. And then x & z where x is the court's length and z is the court's width. I then calculated the angle and velocity of the shot of x with respect to both y and x. I planned to combine the magnitude and angle for x & y with the magnitude and angle for x & z into one vector equation.
This is where I ran into some problems. Can you simply combine two vector equations?

Context: I must develop a vector that models the path of a tennis ball using vectors without physics formulas

I have developed a function that represents the perfect shot in tennis in terms of x & y where x is the court's length and y is the height. And then x & z where x is the court's length and z is the court's width. To do this I calculated the angle and velocity of the shot of x with respect to y and then x with respect to z.

I planned to combine the magnitude and angle for x & y with the magnitude and angle for x & z into one vector equation. I now believe this won't work and am stuck on how to even create a 3d projectile motion vector while making it somewhat practical in the real life after developing a solution to help a tennis player.

I calculated the angle of projection and velocity of x & y by integrating them from acceleration with ax = 0m/s and ay= -9.8m/s to displacement
With the function, I ended up with I created the following vector equation. r_t = (xcos(θ)t)i + (-4.9t^2+xsin(θ)t)j where x is the velocity and theta is the angle of projection
I then let R_x= x cos(θ)t and R_y= -4.9t^2 + xsin(θ)t
R_x was then let equal to the maximum x value the ball can travel through
This was rearranged for t and subbed into R_y
R_y was then let equal to the corresponding y value.
The same process was followed to develop another equation. These equations were plotted into Desmos and the intersect was determined as (16.35, 0.515) therefore it was believed the ball must have a velocity of 16.35m/s and a projection of 0.515 radians to go through said points in the xy plane.
The same process was followed for x & z. where acceleration was integrated to find displacement from ax = 0m/s and az=9m/s (the average velocity of wind for complexity). But the end functions gave no real results where points, were used in the path the ball, must travel. Which made me question my original method.
When calculating for x & z should the velocity and/or the angle of projection of R_x= x cos(θ)t be that found when solving for x & y. hence should the equation be
R_x= 16.35 cos(θ)t
t= r_x/(16.35cos(θ))
R_y= -4.9(r_x/(16.35cos(θ))^2 + xsin(θ)(r_x/(16.35cos(θ))
or
t= r_x/(xcos(θ))
R_y = -4.9(r_x/(xcos(θ))^2 + xsin(θ)(r_x/(xcos(θ))

Please consider:
Values have been modified.
The ball must travel above 4mX at 4.5mZ ( the opposition), and the ball must land on the baseline 25mY on the furthest right-hand side of the court at 9mZ. These points are all with respect to the player, at the origin.
Please mention any misconceptions I have and how I should modify my methodology. P.s typed this out twice as it deleted itself, sorry for any lack of explanation feel free to ask, will reply asap. Keeping generalised as it is for assessment.

 

[BvU]

Hello @themethetion ,

$\qquad$!​

Is this a continuation of your first thread ? In that case I (and the folks that sincerely tried to help you) would really appreciate at least some reaction. Otherwise you get more of the same stuff that may well be way over your head.

without physics formulas

I do wonder why you think you can make do without physics ...

But even math formulas are a lot more legible when typeset properly; I propose you invest the time to learn some $\LaTeX$. well worth it !

$\$

 

[berkeman]

Is this a continuation of your first thread ?

@themethetion -- Please keep the discussion of this schoolwork question in that original thread in the schoolwork forums. This thread is closed.