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anonymous

  • one year ago

NEED HELP ASAP. WILL FAN AND MEDAL. A firecracker shoots up from a hill 150 feet high, with an initial speed of 110 feet per second. Using the formula H(t) = -16t^2 + vt + s, approximately how long will it take the firecracker to hit the ground? Eight seconds Nine seconds 10 seconds 11 seconds

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  1. anonymous
    • one year ago
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    @pooja195 @Concentrationalizing

  2. anonymous
    • one year ago
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    If the hill is 150 feet high, then that's essentially saying that when t = 0, h(t) should equal 150. An initial speed of 110 feet just tells you that v = 110. Using that information, you can rewrite h(t) as a quadratic in t only in which then you would just factor and solve for t.

  3. anonymous
    • one year ago
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    Ok, can you walk me through it? :)

  4. anonymous
    • one year ago
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    Well, if you plug in t = 0, what do you get?

  5. anonymous
    • one year ago
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    h(0) = -16 + s?

  6. anonymous
    • one year ago
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    Well, if t is 0, the -16 goes away also. \(-16(0)^{2} +v(0) + s = s\) . In the end, the problem just tells you that s = 150 and v = 110.

  7. anonymous
    • one year ago
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    Ok :) so what next?

  8. anonymous
    • one year ago
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    Well, that means you have \(h(t) = -16t^{2} +110t + 150\) So, h(t) represents height. You plug in a time value and you get a height for an answer. The problem wants to know when you hit the ground. Well, if you hit the ground, then the height is 0. So we want to know what time value will make h(t) equal 0. Following me so far?

  9. anonymous
    • one year ago
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    Oh ok :D I think so.

  10. anonymous
    • one year ago
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    Mhm. So if h(t) needs to be 0, then you're solving this: \(-16t^{2} + 110t + 150 = 0\)

  11. anonymous
    • one year ago
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    Ok just give me a minute...

  12. anonymous
    • one year ago
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    Alrighty

  13. anonymous
    • one year ago
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    Ok, I think I got lost. I only got to the beginning: -16t^2 + 110t = -150. :(

  14. anonymous
    • one year ago
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    After that I don't know what to do.

  15. anonymous
    • one year ago
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    Well, this is a quadratic equation, so it needs to be factored.

  16. anonymous
    • one year ago
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    Ok, so -2t(8t - 55) = -150?

  17. anonymous
    • one year ago
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    Nope. Have you seen before how an equation like \(x^{2} + 5x + 6\) turns into \((x+2)(x+3)\)?

  18. anonymous
    • one year ago
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    Yes, I have. :) So I need to do that here?

  19. anonymous
    • one year ago
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    Yeah, its that idea. So itll factor and look like ( )( ) = 0 Now it's just knowing how to factor in that way.

  20. anonymous
    • one year ago
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    Hmm. So would it start to look like (-4t +/- ___)(-4t +/- ___)?

  21. anonymous
    • one year ago
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    Actually, this doesnt factor cleanly. Since the question said "approximately", we won't be able to factor in the way I was thinking. We're actually forced to quadratic formula on this, I apologize.

  22. anonymous
    • one year ago
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    Oh ok! That's fine :)

  23. anonymous
    • one year ago
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    Youre okay with doing quadratic formula correctly? :)

  24. anonymous
    • one year ago
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    Yes. Here give me just a few minutes ;) things around here are getting a bit crazy XD

  25. anonymous
    • one year ago
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    Ok so is it -8.04 or 1.17?? Oh so then it would be eight seconds??

  26. anonymous
    • one year ago
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    Sorry I took so long lol

  27. anonymous
    • one year ago
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    I didnt do the full math, but itd be 8 seconds, yeah :3

  28. anonymous
    • one year ago
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    Ok thanks! I'll give you a medal but I'm already a fan ;)

  29. anonymous
    • one year ago
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    Lol, okay. As long as it makes sense.

  30. anonymous
    • one year ago
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    Yes it does now ;) because of you. Thanks so much! Ttyl :)

  31. anonymous
    • one year ago
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    Good luck :3

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