contestada

5. The height, h(t), in feet, of an object shot from a cannon with initial velocity of 20 feet per second can
Wbe modeled by the equation h(t)=-16t2 +20t+6, where r is the time, in seconds, after the cannon is
fired. What is the maximum altitude that the object reaches?

Respuesta :

msm555

Answer:

12.25 feet

Step-by-step explanation:

The maximum altitude of the object can be found by identifying the vertex of the quadratic equation [tex] h(t) = -16t^2 + 20t + 6 [/tex].

The vertex form of a quadratic equation is given by:

[tex] f(x) = a(x - h)^2 + k [/tex],

where

  • [tex](h, k)[/tex] is the vertex.

In this case, [tex] a = -16 [/tex], so we need to rewrite the equation in vertex form.

[tex] h(t) = -16t^2 + 20t + 6 [/tex]

To find the vertex, we can use the formula [tex] h = -\dfrac{b}{2a} [/tex].

In this equation, [tex] b [/tex] is the coefficient of the linear term, and [tex] a [/tex] is the coefficient of the quadratic term.

[tex] h = -\dfrac{20}{2(-16)} [/tex]

[tex] h = -\dfrac{20}{-32} [/tex]

[tex] h = \dfrac{5}{8} [/tex]

Now, substitute [tex] t = \dfrac{5}{8} [/tex] into the original equation to find the maximum altitude ([tex] k [/tex]):

[tex] k = -16\left(\dfrac{5}{8}\right)^2 + 20\left(\dfrac{5}{8}\right) + 6 [/tex]

Calculate the value of [tex] k [/tex] to find the maximum altitude.

[tex] k = -16\left(\dfrac{25}{64}\right) + \dfrac{100}{8} + 6 [/tex]

[tex] k = -\dfrac{400}{64} + \dfrac{800}{64} + \dfrac{384}{64} [/tex]

[tex] k = \dfrac{-400 + 800 + 384}{64} [/tex]

[tex] k = \dfrac{784}{64} [/tex]

[tex] k = 12.25 [/tex]

Therefore, the maximum altitude that the object reaches is 12.25 feet.