Question

An object is launched directly in the air speed of 16 feet per second from a platform located 5 feet above the ground. The position of the object can be modeled using the function f(x)=-16t^2+16t+5, where t is the time of seconds and f(t) is the height of the object. What is the maximum height in feet that the object will reach?

Answer 1

24

Explanation:

Answer 2

The maximum height that the object will reach is of 9 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:

`f(t) = -16t^(2) + 16t + 5`

So

`a = -16, b = 16`

The instant of the maximum height is:

`t_(v) = -(16)/(2*(-16)) = 0.5`

The maximum height is:

`f(0.5) = -16*(0.5)^2 + 16*0.5 + 5 = 9`

The maximum height that the object will reach is of 9 feet.