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ladiesman217

  • one year ago

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  1. ladiesman217
    • one year ago
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    If a baseball player hits a baseball from 4 feet off the ground with an initial velocity of 64 feet per second, how long will it take the baseball to hit the ground? Use the equation \[h=-16t ^{2}+64t+4\]

  2. mukushla
    • one year ago
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    baseball hits the ground when \(h=0\).

  3. ladiesman217
    • one year ago
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    ok

  4. ladiesman217
    • one year ago
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    @Shadowys @UnkleRhaukus @TSwizzle can you help me please

  5. Shadowys
    • one year ago
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    \(h=-16t^2 +64t +4\) when h=0,(the ball hits the ground) \(0=-16t^2 +64t +4\) use the general formula. what did u get?

  6. sami-21
    • one year ago
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    you need to solve the quadratic equation since h=0 so -16t^+64t+4=0 or 16t^2-64t-4=0 use quadratic formula with a=16 b=-64 c=-4 you will get two roots t=4 sec t=-0.6 but time cannot be negative so neglect the negative value t=4 sec is the required time .

  7. ladiesman217
    • one year ago
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    \[\frac{ -64\sqrt{64^{2}+16*-4} }{ 2*16 }\] \[\frac{ -64\sqrt{64^{2}-64} }{ 32 }\]

  8. sami-21
    • one year ago
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    it should be +64 outside the root because b=-64 and there is minus in the formula too .

  9. Shadowys
    • one year ago
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    note: it's \(-64 \pm \sqrt{64^2 - (4)(-16)(4)}\) up there.

  10. sami-21
    • one year ago
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    correct formula is \[\Large =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  11. ladiesman217
    • one year ago
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    \[\frac{ 64\sqrt{64^{2}+256} }{32}\]

  12. sami-21
    • one year ago
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    you miseed plus minus sign :)

  13. Shadowys
    • one year ago
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    yup! simplify and you're done (i guess the editor don't have it. the latex is \pm

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