Respuesta :

fieryanswererft

Answer:

(a) The maximum height of the ball is 16 feet at 1 seconds.

(b) The ball hits the ground at 2 seconds.

PART (A) :

Given function: h(t) = -16t² + 32t

Comparing to quadratic function: ax² + bx + c

In this function: a = -16, b = 32, c = 0

To find the maximum height, use the vertex formula:

[tex]\sf x = \dfrac{-b}{2a}[/tex]

Insert values

[tex]\rightarrow \sf t = \dfrac{-32}{2(-16)} = 1[/tex]

Then find h(t) = -16(1)² + 32(1) = 16 feet

Conclusion: The maximum height of the ball is 16 feet at 1 seconds.

PART (B) :

When the ball hits the ground, the height [h(t)] will be 0 ft

-16t² + 32t = 0

-16t(t - 2) = 0

-16t = 0, t - 2 = 0

t = 0, t = 2

The ball hits the ground at 2 seconds. Note: the other 0 seconds is for when the ball was launched at the beginning.

#a

Find t coordinate of vertex

  • -b/2a
  • -32/2(-16).
  • -32/-32
  • 1

Put in equation

[tex]\\ \rm\Rrightarrow h(1)=-16+32[/tex]

[tex]\\ \rm\Rrightarrow H_{max}=16ft[/tex]

#b

  • Make h(t)=0

[tex]\\ \rm\Rrightarrow -16t^2+32t=0[/tex]

[tex]\\ \rm\Rrightarrow -16t(t-2)=0[/tex]

[tex]\\ \rm\Rrightarrow t=0,2[/tex]

  • At 0s and 2s the ball will be at ground

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