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Math Help - Word problems

  1. #1
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    Word problems

    A model rocket is launched from the roof of a building. It's flight path through the air is modeled by h=-5tē+23t+10, where h is the height of the rocket above ground in meters and t is the time after the launch in seconds.

    A) when does the rocket hit the ground?

    I got this far H= -1(5tē-23t-10) -23-50
    H= -1(t-25)(t+2) -25+2
    The teacher answer has this..........
    H= -1(5tē-23t-10)
    H= -(t-5)(5t+2)

    t-5= 0 5t+2=0
    t=5 t=-2/5

    Where does 5t come from?
    I don't get the whole H= -5tē+23t+10 equation what does -5 stand for? 23t? 10?
    Teacher just taught me the equation and I follow it but, I kinda dont get the whole thing.
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  2. #2
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    How did you get these steps?

    Quote Originally Posted by buck View Post
    I got this far H= -1(5tē-23t-10) -23-50
    H= -1(t-25)(t+2) -25+2
    All you need to do is solve where  h=0

    i.e. 0=-5tē+23t+10

    You can factor the RHS and apply the null factor law like in your teachers solution or just use the quadratic formula.

    Which method do you prefer?
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  3. #3
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    I don't know my teachers method. I don't know which method I use but if I bring down t-25=0 wouldn't 5 be 25?
    My teacher got a totally different answer because of the (5t+2)
    i dont know how she got that
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  4. #4
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    Quote Originally Posted by buck View Post
    I don't know which method I use but if I bring down t-25=0 wouldn't 5 be 25?
    you may have to clarify further what you mean by this.

    I would solve using the quadratic formula \displaystyle t=\frac{-23\pm\sqrt{23^2-4\times -5\times 10}}{2\times  -5}
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