zaanebali1
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A ball is thrown from a height of 44 meters with an initial downward velocity of 6 m/s . The ball's height h (in meters) after t seconds is given by the following.
h= 44 - 6t - 5t squared 2
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Sagot :

rspill6

Answer:

Step-by-step explanation:

h = -5t^2 - 6t + 44

I'll assume 0 meters is the ground height.  That would mean we want to know the time, t, for h to be = 0 meters:

h = -5t^2 - 6t + 44

0 = -5t^2 - 6t + 44

5t^2 + 6t -44 = 0

Solve using the quadratic equation:  2.43 and - 3.63 seconds.  We'll use the positive value:  2.43 seconds for the ball to reach the ground.  Save the -3.63 value for the Klingons.  

We can also solve by graphing the function, as per the attached image.  Note that the starting time of 0 seconds, the ball is at 44 feet.  It reaches the x axis at 2.43 seconds (where x = 0, the ground).

View image rspill6
semsee45

Answer:

t = 2.43 s (nearest hundredth)

Step-by-step explanation:

Given equation: [tex]h=44-6t-5t^2[/tex]

where:

  • h = height (in meters)
  • t = time (in seconds)

When the ball hits the ground, h = 0

[tex]\implies 44-6t-5t^2=0[/tex]

[tex]\implies 5t^2+6t-44=0[/tex]

Using the quadratic formula when [tex]ax^2+bx+c=0[/tex] and where:

  • a = 5
  • b = 6
  • c = -44
  • x = t

[tex]t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

  [tex]=\dfrac{-(6)\pm\sqrt{(6)^2-4(5)(-44)}}{2(5)}[/tex]

  [tex]=\dfrac{-6\pm\sqrt{916}}{10}[/tex]

  [tex]=\dfrac{-6\pm2\sqrt{229}}{10}[/tex]

  [tex]=\dfrac{-3\pm\sqrt{229}}{5}[/tex]

  [tex]=2.43, -3.63 \ \textsf{(nearest hundredth)}[/tex]

As time is positive, t = 2.43 s (nearest hundredth)

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