Projectile Motion Using Vectors

[themethetion]

It avoids the complication of the launch angle needing to change when you introduce the lateral motion,

Yes, and of the opponent’s position.

But the opponent is on the diagonal, so wouldn't his position remain the same?

 

[haruspex]

But the opponent is on the diagonal, so wouldn't his position remain the same?

The position on the court is the same, but not the distance from the origin.

 

[themethetion]

The position on the court is the same, but not the distance from the origin.

But the origin hasn't changed?

 

[haruspex]

But the origin hasn't changed?

In your method in post #26, you started off by ignoring the width of the court. You treated the trajectory as being straight up one side of the court, baseline to baseline, with the opponent half way.
Changing it to a corner to corner trajectory increases the distance from the server to the opponent as well as from server to landing point.

 

[pbuk]

I am sure that you have the skills to work this out if you think clearly and work methodically, but after 34 posts you still seem to be struggling. I think there are two main reasons for the struggle. First you don't seem to have a mental image of the problem. The best way to get a mental image is to draw a diagram: do this and post it here, with everything labelled. Which brings me to the second problem: you have used non-standard and very confusing labels for things, in particular

x is the velocity

Don't do this. Choose $v$ for the initial velocity (actually it is speed, but let's stick with $v$ because we all know we are talking about the magnitude of the velocity here), then you can use $x$ instead of R_x and $y$ instead of R_y. Now using $g$ for gravity (always keep symbols in your workings, only substituting numbers in at the end to get an answer) we can rewrite

R_x= x cos(θ)t and R_y= -4.9t^2 + xsin(θ)t

as $x = (v \cos \theta) t$ and $y = y_0 + (v \sin \theta) t - (\frac 1 2 g) t^2$ (note I have also introduced $y_0$, the initial height of the ball as I am sure the player is not hitting it off the ground). We can then use sensible labels for things like the distance to the baseline $x_b$ and to the distance and height to clear the other player $x_p$ and $y_p$. Mark these on your diagram.

 

[FranzS]

Is this some math exercise or do you want it to be of any actual use for tennis practice? Drag and Magnus effect due to ball spin are not negligible in the real world.

 

[PeroK]

Is this some math exercise or do you want it to be of any actual use for tennis practice? Drag and Magnus effect due to ball spin are not negligible in the real world.

Not to mention that a lob may be hit from well behind the baseline and that most players don't remain statuesque at the net, but run backwards to hit a smash.

 

[kuruman]

... most players don't remain statuesque at the net ...

... with their racquets raised up high like a present-day Statue of Liberty ...

 

[themethetion]

Is this some math exercise or do you want it to be of any actual use for tennis practice? Drag and Magnus effect due to ball spin are not negligible in the real world.

Not to mention that a lob may be hit from well behind the baseline and that most players don't remain statuesque at the net, but run backwards to hit a smash.

It's a math exercise that we must make as practical as possible to 'help inform players in performing a successful shot'.

 

[neilparker62]

TL;DR Summary: Using vector functions how can I find the minimum average velocity (something greater than 11.86m/s) of a ball if the launch angle is unknown and if I have a point that the object must travel through (11.86, 3.47)?

Not sure about vector "functions" as such but you can use:

$$ \vec{a}\times\vec{s} = \vec{v} \times \vec {u} $$

As @kuruman has elsewhere advised, minimum velocity corresponds to $\vec{v}$ and $\vec{u}$ being perpendicular. In this case, the above simplifies to:

$$gR = u\sqrt{u^2-2gh}$$ where R is the given x-coordinate and h the given y-coordinate.

Edit y-coordinate should be less one given the ball is hit from 1m above ground. See post #4.

https://www.desmos.com/calculator/3byo9fl9ha