Now let's see if you numpties can understand conceptual ideas and the proper use of maths.
I understand it just fine. I defined u to be the (initial) speed of the ball. if it's easier, just treat it as x. It's fair enough as applying SUVAT to the Earth, u=0, so there's little risk of confusion for anyone who more than skims my calculations. I'm just more comfortable using unknowns, it's often clearer what's going on.
You seem to be relying on a flawed intuitive understanding. When you throw a ball, the Earth reaches the ball's speed: this doesn't mean the Earth is halfway to the ball.
You're also calculating the time it takes for the Earth to go 10m (or whatever units you prefer) into the air. That doesn't make sense, unless you're defining that to be the point at which the ball begins to descend/the Earth matches the speed of the ball, but that really doesn't seem to be what you're doing. You're confusing the two objects under discussion; the ball moves 10m into the air (in the reference frame of the Earth), but the Earth moving 10m has no significance. The ball is always going to be moving up with its starting speed (treating air resistance as negligible) so it would in fact be travelling more than 10m. Yes, you said this, but you ignored its relevance. What is the benefit of calculating the time it takes for the Earth to move 10m, when that has no relevance to the distance the ball moves, or the relative velocities of the ball and Earth, or indeed anything that's actually under discussion?
To rewrite my previous calculations with a letter changed as that apparently confused you:
I await to hear what flaw you find.
So, let's throw a ball up with fixed velocity x>0, where we model the Earth's starting velocity to be zero.
How long is it before the Earth's velocity is x? In practise, this will be modelled by the top of the ball's arc, before it seems to begin descending. (It's the point at which the Earth begins to move faster than the ball, so the ball looks like it's falling).
Well, a=9.8, u=0, v=x, we want to find t.
Use the first SUVAT equation v=u+at to find t = x/9.8
Now then, at this point, the Earth and the ball are travelling at the same speed, so it will seem as though the ball is stationary in the air. A moment before, it was ascending: a moment after, it's descending.
Let's see what happens when t = 2u/9.8. This is twice the time taken for the ball to reach the apex of its flight. We use the second equation, your s=ut + 0.5at
2The displacement of the ball (u=x, a=0) is s=ut = 2x
2/9.8
The displacement of the Earth (u=0, a=9.8 ), using the same equation, is s=0.5at
2 = 2x
2/9.8
That is, the ball is back to its starting point, and it takes the same time to go up as it does to go down.
Trust me, I understand maths. First class BSc degree, and about a month away from an MSc, and happy to walk you through my dissertation if you've any interest in ratios of eigenvalues of Schrodinger operators. This topic, however, is trivially simple.