Question

The function h(t) --4.922 +17.69+575 is used to model the height of an object being tossed from a tall building, where h(t) is

Answer 1

Answer:

Domain: [0, 12.75]

Range: [0, 590.9]

Step-by-step explanation:

Your function is missing variables.

I think you mean

h(t) = -4.922t^2 + 17.69t + 575

as the function.

We can find the time the object hits the ground. At that time, h(t) = 0, so we set the equation equal to zero and solve for t.

-4.922t^2 + 17.69t + 575 = 0

$t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t = \dfrac{-17.69 \pm \sqrt{(-17.69)^2 - 4(-4.922)(575)}}{2(-4.922)}$

$t = \dfrac{-17.69 \pm \sqrt{11633.54}}{-9.844}$

t = -9.16 s or t = 12.75 s

According to the parabola that is the path of the falling object, the object is at zero height at t = -9.16 s and t = 12.75 s. Since the object starts moving at time t = 0, the domain has to be limited to t = 0 till t = 12.75 s.

Domain: [0, 12.75]

The range is the height of the object. Maximum height occurs at the time that is the midpoint of the two times the parabola shows h(t) = 0.

tmax = (-9.16 + 12.75)/2 = 1.797

Now we use the time of maximum height to find the maximum height.

h(t) = -4.922t^2 + 17.69t + 575

h(tmax) = -4.922(1.797)^2 + 17.69(1.797) + 575

h(tmax) = 590.9

The maximum height is 590.9 m.

Range: [0, 590.9]