Question

PLEASE HELP ASAP

A ball is thrown straight up from the top of a building 144 ft tall with an initial velocity of 48 ft per second. The height
s(t) (in feet)
of the ball from the ground, at time t (in seconds), is given by
s(t) = 144 + 48t − 16t^2.
Find the maximum height attained by the ball.

Answer 1

The maximum height attained by the ball is of 180 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:

We have that:

`s(t) = -16t^2 + 48t + 144`

Which is a quadratic equation with
`a = -16, b = 48, c = 144`.

The maximum height is the value of s, which is the output, at the vertex. So

`\Delta = b^2-4ac = 48^2 - 4(-16)(144) = 11520`

`s_(v) = -(11520)/(4(-16)) = (11520)/(64) = 180`

The maximum height attained by the ball is of 180 feet.