Aball is thrown from 8 feet off the ground with a vertical velocity of 28 feet per second. is the ball at its maximum height after 1 second? explain.

No. The maximum height occurs earlier.

The equation for ballistic motion is typically written as
  height = -(1/2)g·t² +v₀·t +h₀
where v₀ and h₀ are the initial vertical speed and height, respectively. The constant "g" is typically considered to be 32 ft/s² in algebra problems.

Thus your motion equation can be written as
  height = -16t² + 28t + 8
The vertex of quadratic ax²+bx+c is found at x=-b/(2a). The vertex of this quadratic will be located at
  t = -28/(2·(-16)) = 7/8
This is less than 1 second, so the ball is already on its way down at t=1.

2,4,5,8,10 are divisible by 40