Question

You've been asked to estimate the maximum height of a ball thrown from ground level. The equation that represents the height of
the ball as a function of time (t) in seconds is: h = -16t² + 54.4t
What is the realistic range for this height scenario?

Answer 1

Using the vertex of the quadratic equation, a realistic range for this height scenario is given by: [0, 46.24].

What is the vertex of a quadratic equation?

A quadratic equation is modeled by:

y = ax^2 + bx + c

The vertex is given by:

$(x_v, y_v)$

In which:

• $x_v = -\frac{b}{2a}$

• $y_v = -\frac{b^2 - 4ac}{4a}$

Considering the coefficient a, we have that:

• If a < 0, the vertex is a maximum point.

• If a > 0, the vertex is a minimum point.

In this problem, the ball starts at the ground, hits it's maximum height than falls to the ground again, hence a reasonable range is between 0 and the maximum height.

The equation is:

h = -16t² + 54.4t.

The coefficients are a = -16, b = 54.4, c = 0, hence the maximum height is:

$y_v = -\frac{54.4^2 - 4(-16)(0)}{4(-16)} = 46.24$

A realistic range for this height scenario is given by: [0, 46.24].

More can be learned about the range of a quadratic equation at https://brainly.com/question/24737967

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