I completely understand what you're saying. Let me explain it in a different way...
Suppose i just dropped that ball right? The acceleration is obviously g downwards. So the ball falls towards the ground, accelerating at 9.81 (metres per second) per second. The only acceleration acting downwards is this, and therefore you can use SUVATs to find time.
Now lets consider that i throw the ball, from the same height. The acceleration acting down on the ball is
still only g. This means that the ball, as before, is accelerating towards the ground at 9.81 (metres per second) per second. Therefore the only acceleration acting downwards is this, like the dropping scenario, as they are both dropped from the same height they both hit the ground
at the same timeNow i hear you saying in your head,
oh but does the one you throw not travel further and so must be in the air for longer?No
The balls both hit the ground when their vertical distance travelled is 24m right?
If they are both accelerating towards the ground at 9.81ms^-2, then they both reach the ground at the same time.
The horizontal velocity of the ball is not a factor. Im going to use some diagrams to show this effect visually, where i show the distance travelled in a time.
This is me dropping the ball from a height right? now lets compare to if i threw that ball:
Can you see how they travel the same vertical distance in the same time?
This shows that the time spent in the air is the same in both cases from the same height. Hope this helped.
Edit:
You're not helping. This stuff is not 'easy'. Anyway this is the maths science and technology academic
help section.
If you're not going to help, GTFO.