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

Answer:

The ball reaches maximum height in 3 seconds.

The maximum height of the ball is of 400 feet.

Step-by-step 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

$x_{v} = -\frac{b}{2a}$

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:

$t_{v} = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3$

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.