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help! h(t)=-16t^2 +128 t
An arrow is fired into the air with an initial velocity of 128 feet per second. The height in feet of the arrow t seconds after it was shot into the air is given by the function above. h(t)=-16t^2 +128 t After how many seconds does the arrow take to reach its maximum height?


a)4 seconds

b).25 seconds

c)8 seconds

d)3 seconds


In the previous problem, you determined the time it takes for the arrow to reach its maximum height. Use that information to find the maximum height of the arrow.


a)256 ft

b)896 ft

c)240 ft

d)31 ft

Respuesta :

mysticchacha

Answer:

It takes 4 seconds

Step-by-step explanation:

Need to find the Vertex of h(t)

to find maximum height.

Vertex at  t = -b/(2a)

t = -(128) / (2*-16) =  4 seconds

      Answer for part (1) of the question is Option (a) and for part (2) is Option (a).

Function representing the height of the arrow has been given as,

  • h(t) = -16t² + 128t

        Here, h(t) = Height of the arrow after time 't' seconds

                      t = time

 Given function is a quadratic function, maximum point will be its vertex represented by the coordinates → [tex][-\frac{b}{2a}, h(-\frac{b}{2a})}][/tex]

By comparing the given equation of the function with the quadratic function,

f(x) = ax² + bx + c

a = -16, b = 128, c = 0

To find the coordinates of the vertex,

[tex]\frac{b}{2a}=\frac{128}{-2(16)}[/tex]

    [tex]=-4[/tex]

[tex]h(-\frac{b}{2a})=h(4)=-16(4)^2+128(4)[/tex]

                       [tex]=-256+512[/tex]

                       [tex]=256[/tex] feet

Coordinates of the vertex → (4, 256)

Here, x-coordinate represent the time and y-coordinate represents the maximum height.

Therefore, at t = 4 seconds arrow will be at maximum height 256 feet.

    Hence, answer for part (1) of the question will be Option (a) and for part (2) will be Option (a).

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