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Hatshy [7]
3 years ago
7

A ball is thrown into the air with an upward velocity of 44 ft/s. It’s height h in feet after t seconds is given by the function

h=-16t^2 +44t+5. How long does it take the ball to reach its maximum height? What is the balls maximum height?
Mathematics
1 answer:
MAVERICK [17]3 years ago
5 0

9514 1404 393

Answer:

  • 1.375 seconds
  • 35.25 feet

Step-by-step explanation:

A quadratic function y=ax^2 +bx +c is symmetric about the vertical line x=-b/(2a). That is the x-coordinate of the vertex (maximum or minimum).

Here, we have ...

  h = -16t^2 +44t +5

  t = -44/(2(-16)) = 11/8 = 1.375

It takes 1.375 seconds for the ball to reach its maximum height.

The height at that time is ...

  h = (-16(11/8) +44)(11/8) +5 = (-22 +44)(11/8) +5 = 121/4 +5 = 141/4 = 35.25

The maximum height is 35.25 feet.

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\frac{1}{2}  which agrees with answer B

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Is 7/8 or 11/10 closer to 1? How did you decide anybody help please?
erik [133]

Let

A-----> \frac{7}{8}

B-----> \frac{11}{10}

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we know that

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\frac{7*10}{8*10} =\frac{70}{80}

Point B-------> Multiply numerator and denominator by 8

\frac{11*8}{10*8} =\frac{88}{80}

Point C-------> Multiply numerator and denominator by 80

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the answer is

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