The ball's instantaneous acceleration at t = 1 is given by

              
             d^2 y
      a(t) = -----   =  - 9.8 m/s^2
             dt^2


   To find the ball's average acceleration between t = 0 and t = 1,
we need first to compute the ball's velocity at those times:

     
     v(t)  =  3 (m/s)  -  9.8 (m/s^2) * t

so

     v(0)  =  3 m/s

     v(1)  =  3 m/s  -  9.8 (m/s^2) * 1 s  =  -6.8 m/s


And so the average acceleration is

                     -6.8 m/s   -   3 m/s
     avg accel  =   ---------------------   =   -9.8 m/s^2
                       1 s    -    0 s


  Curious.  The average acceleration is equal to the instantaneous
acceleration ...