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help_people

  • one year ago

Amir pitches a baseball at an initial height of 6 feet, with a velocity of 73 feet per second. If the batter misses, about how long does it take the ball hit the ground? Hint: Use H(t) = −16t2 + vt + s. 4.64 seconds 2.94 seconds 2.28 seconds 0.08 second

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  1. welshfella
    • one year ago
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    when the ball hits the ground H(t) = 0

  2. welshfella
    • one year ago
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    so plug in 0 for H(t) , 6 for s and 73 for v, then solve for t.

  3. help_people
    • one year ago
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    h(0)=-16t^2+73t+6? @welshfella

  4. help_people
    • one year ago
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    then i would solve for t which i will do in a sec

  5. help_people
    • one year ago
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    t=250/73?

  6. welshfella
    • one year ago
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    i got a different value

  7. welshfella
    • one year ago
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    250/73 = 3.42

  8. welshfella
    • one year ago
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    did you use the quadratic formula?

  9. welshfella
    • one year ago
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    x = [ -73 +/- sqrt(73^2 - 4*-16*6) ] / (2*-16)

  10. help_people
    • one year ago
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    oh ok its not different you just divided lol

  11. help_people
    • one year ago
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    what would i do after?

  12. welshfella
    • one year ago
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    = (-73 - 75.58) / -32 = 4.64

  13. welshfella
    • one year ago
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    73^2 - 4*-16*6 comes to 5713 and sqrt 5713 = 75.58

  14. welshfella
    • one year ago
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    - the other root is 0.08 which we can ignore

  15. help_people
    • one year ago
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    ok

  16. welshfella
    • one year ago
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    - sorry it was -0.08 which of course we can ignore

  17. help_people
    • one year ago
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    ok so the answer would be 4.64? based on what you have explained/said

  18. welshfella
    • one year ago
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    yes

  19. welshfella
    • one year ago
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    the other root is negative - the correct time is 4.64 seconds

  20. help_people
    • one year ago
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    ty! :)

  21. welshfella
    • one year ago
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    yw

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spraguer (Moderator)
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