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A ball is thrown from an initial height of 3 feet with an initial upward velocity of 44 ft/s. The ball's height h (in feet) after seconds is given by the following.
h = 3+44t-16t^2

Find all values of t which the balls height is 23 feet.

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Answer:

See below ↓↓

Step-by-step explanation:

Given function

  • h = 3 + 44t - 16t²

We need to calculate the possible values of t at h = 23 feet.

Subsituting h = 23,

  • 23 = 3 + 44t - 16t²
  • 16t² - 44t + 20 = 0
  • Divide throughout by 4 as it is a common factor
  • 4t² - 11t + 5 = 0

Solving for 't' using quadratic formula

  • t = 11 ± √121 - 4(4)(5) / 8
  • t = 11 ± √41 / 8

Solutions

  1. t = 11 + 6.4 / 8 = 17.4/8 = 2.175 seconds
  2. t = 11 - 6.4 / 8 = 4.6/8 = 0.575 seconds

semsee45

Answer:

t = 2.175 s (3 dp)

t = 0.575 s (3 dp)

Step-by-step explanation:

Given equation: [tex]h=3+44t-16t^2[/tex]

To find all values of t for which the ball's height is 23 ft, substitute h = 20 into the equation and solve for t:

[tex]\implies h=23[/tex]

[tex]\implies 3+44t-16t^2=23[/tex]

[tex]\implies 16t^2-44t-3+23=0[/tex]

[tex]\implies 16t^2-44t+20=0[/tex]

Factor out common term 4:

[tex]\implies 4(4t^2-11t+5)=0[/tex]

Divide both sides by 4:

[tex]\implies 4t^2-11t+5=0[/tex]

Quadratic formula

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]

Use the quadratic formula to solve for t:

[tex]\implies t=\dfrac{-(-11) \pm \sqrt{(-11)^2-4(4)(5)} }{2(4)}[/tex]

[tex]\implies t=\dfrac{11 \pm \sqrt{41}}{8}[/tex]

Therefore,

  • t = 2.175 s (3 dp)
  • t = 0.575 s (3 dp)

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