a. When t=0, H(0)=2meters,and that is the height when the ball is released. b. To caculate the max height, write down the derived function:H'(t)=-10t+6; and let the derived function:H'(t)=-10t+6 to be zero , H'(t)=-10t+6=0 , so t=0.6; H(0.6)=-5(0.6)^2+6*(0.6)+2=3.8metes; c. H(t)=-5t^2+6t+2=3, t=0.2or1, "on its way down" , so t=1 is the right answer.