Question

1. A baseball is thrown directly upward with

an initial velocity of 96 feet per second
from an initial height of 7 feet. The height
of the ball (in feet) after t seconds is given
by the formula h(t) = -16 t + 96t + 7.
a) After how many seconds will the ball
reach its maximum height?
b) What is that maximum height?

Answer 1

a) The ball reaches it's maximum height after 3 seconds.

b) The maximum height of the ball is of 151 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), y_(v))`

In which

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

`y_(v) = -(\Delta)/(4a)`

Where

`\Delta = b^2-4ac`

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

In this question:

The height of the ball is modeled by:

`h(t) = -16t^2 + 96t + 7`

So a quadratic equation with
`a = -16, b = 96, c = 7`

a) After how many seconds will the ball reach its maximum height?

t-value of the vertex. So

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

The ball reaches it's maximum height after 3 seconds.

b) What is that maximum height?

h of the vertex.

`\Delta = b^2 - 4ac = (96)^2 - 4(-16)(7) = 9664`

`h_(v) = -(9664)/(4(-16)) = 604`

The maximum height of the ball is of 151 feet.