The formula for the height of a ball as a function of time is given by the equation h = - 16t^2 + vt + h, where h is the height of a ball in feet, v is the initial velocity of the ball in feet per second, h is the initial height of the ball in feet, and t is the time in seconds after the ball was thrown.

If a ball is thrown from an initial height of 5 feet at an initial velocity of 20 feet per second, what is its height after 1 second?

It's really poor form to use 'h' to mean two different things in two different

places in the same formula. Fortunately, we know what you mean.

H = h + vt - 16t²

You have said that ...

initial height = 5 feet

initial velocity = 20 feet per second, no direction given

time = 1 second

Since you gave us the magnitude of the initial velocity but not its direction,

there are a huge number of possibilities. I'll focus only on two of them:

20 feet per second straight down, and 20 feet per second straight up.

Initial velocity is downward:

H = (5) + (-20) (1) - 16 (1) ²

H = 5 - 20 - 16 = 5 - 36 = 31 feet underground after 1 second

Initial velocity is upward:

H = (5) + (20) (1) - 16 (1) ²

H = 5 + 20 - 16 = 25 - 16 = 9 feet above ground after 1 second