May day may day,
i have an urgent problem, 14 points to the answer-er!
adam tossed a baseball into the air with an initial upward velocity of 10 meters per second.
a. write an equation relating the distance in meters of the ball above where it was tossed t seconds after it was tossed
use dis d=rt - 5t^2
b. how high above where it was tossed would the ball be after 0.5

Have you heard the expressions "shooting fish in a barrel" or
"taking candy from a baby" ?

I've decided to take your points and try not to feel guilty about it.

A).  That's it !  That's the equation, for ANY basketball on Earth. 
You wrote it right there in your question.

In that equation that you wrote, the ' r ' is the initial velocity. 
You said that Adam tossed it straight up, and it was going 10 meters per second
   when it left his hand. 
So ' r ' in the equation is +10.
The equation is:
                             d = 10t - 5t²    .
It tells you how high the ball is above its tossing height after any number of seconds.
The ' t ' in the equation is the number of seconds.  Any letter could have been used,
but ' t ' was cleverly selected because it stands for 'time'.

B).  You want to know where it is after 0.5 second ?    All you have to do is put '0.5'
into the equation wherever there's a ' t '.  Do you really need somebody to do that
for you ?

Well, OK.  You're being so overly generous with your points . . .

         Distance = 10 t - 5 t²

         Distance = 10 (0.5) - 5 (0.5)²

         Distance =     5       - 5(0.25)

         Distance =     5       -  1.25

         Distance =        3.75 meters

                                positive 3.75 means above Adam's hand. 

A thinking exercise:
-- He tossed the ball upwards at  10  meters per second.
-- Why is it not  5  meters above his hands after  0.5  second?

The correct answer is -3.9> -4.5 gucci playa

So confused right here i need it to be more clearer