A ball 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.