I think you mean to have -4.912t^2 in there. So if you take the derivative of the function, you get df/dt = -9.824t + 28.62. The df/dt is 0 when finding a max or min. So 0 = -9.824t + 28.62, and subtracting both sides by 28.62 then dividing the resulting equation by -9.824 results in the max height being at approximately 2.913 seconds. So an appropriate window would be 2.5 to 3.5 seconds.