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:

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

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

[tex]f(x) = ax^{2} + bx + c[/tex]

It's vertex is the point [tex](x_{v}, y_{v})[/tex]

In which

[tex]x_{v} = -\frac{b}{2a}[/tex]

[tex]y_{v} = -\frac{\Delta}{4a}[/tex]

Where

[tex]\Delta = b^2-4ac[/tex]

If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].

In this question:

We have that:

[tex]s(t) = -16t^2 + 48t + 144[/tex]

Which is a quadratic equation with [tex]a = -16, b = 48, c = 144[/tex].

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

[tex]\Delta = b^2-4ac = 48^2 - 4(-16)(144) = 11520[/tex]

[tex]s_{v} = -\frac{11520}{4(-16)} = \frac{11520}{64} = 180[/tex]

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

PLEASE ANSWER ASAPPPP

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