Question

An object is launched directly in the air at a speed of 64 feet per second from a platform located 16 feet above the ground. The position of the object can be modeled using the function f(t)=−16t2+64t+16, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

Answer 1

Considering the vertex of the quadratic equation, the maximum height that the object will reach is of 80 feet.

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 equation is:

f(t) = -16t² + 64t + 16.

Hence the coefficients are:

a = -16, b = 64, c = 16.

The maximum value is found as follows:

$y_v = -\frac{64^2 - 4(-16)(16)}{4(-16)} = 80$

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

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