Question

S(t) = -16t^2+96t+256 the quadratic function models the balls height above the ground, s(t), in feet, t seconds after it was thrown when a person standing close to the edge on top of a 248-foot building throws a baseball ball vertically upward. When will the ball reach its maximum height, and what is the maximum height of the ball?

Answer 1

The ball reaches maximum height in 3 seconds.

The maximum height of the ball is of 400 feet.

Explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

`f(x) = ax^(2) + bx + c`

It's vertex is the point
`(x_(v), f(x_(v))`

In which

`x_(v) = -(b)/(2a)`

If a<0, the vertex is a maximum point, that is, the maximum value happens at
`x_(v)`, and it's value is
`f(x_(v))`

In this question:

The height is modeled by:

`s(t) = -16t^2 + 96t + 256`

So, the coefficients are:

`a = -16, b = 96, c = 256`

Instant of time the ball reaches maximum height:

`t_(v) = -(96)/(2(-16)) = -(96)/(-32) = 3`

The ball reaches maximum height in 3 seconds.

What is the maximum height of the ball?

This is s(3).

`s(3) = -16t^2 + 96t + 256 = -16*3^2 + 96*3 + 256 = 400`

The maximum height of the ball is of 400 feet.