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Garcia_123

  • 3 years ago

*PLEASE HELP*!!COIN IS TOSSED UPWARD FROM A BALCONY 188FT HIGH WITH AN INITIAL VELOCITY OF 32FT PER SEC. THE HEIGHT OF THE COIN, IN FT AFTER "T" SECONDS IS GIVEN BY FUNCTION h(t)=-16^2+32+188. FOR WHAT LENGTH OF TIME WILL THE COIN BE AT THE HEIGHT OF AT LEAST 60 FT?

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  1. vacationeer
    • 3 years ago
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    where are the "T's" in the equation?

  2. vacationeer
    • 3 years ago
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    cuz right now with what you have... it would just be.... 476 no matter what... which is impossible

  3. Garcia_123
    • 3 years ago
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    That's all the info i have. I have the answer which is 0 sec to 4 sec but not sure how they got that.

  4. vacationeer
    • 3 years ago
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    O.o.......i mean the equation would work better if it was lik \[h(t)=-16(T^{2})+32T+188\]

  5. mandonut
    • 3 years ago
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    find time at max height

  6. mandonut
    • 3 years ago
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    take the first dervivative

  7. vacationeer
    • 3 years ago
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    that would make more sense... in which you would substitute h(t) for 60 and then solve from there

  8. vacationeer
    • 3 years ago
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    @mandonut ..... its not a max question... its asking for the position of the ball

  9. vacationeer
    • 3 years ago
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    *coin

  10. vacationeer
    • 3 years ago
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    so then it would be \[60=-16(T^{2})+32T+188\]

  11. vacationeer
    • 3 years ago
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    pull the 60 over to get.... \[0=-16(T^{2}) +32T +128\]

  12. vacationeer
    • 3 years ago
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    now you can take out 16 from the whole equation to get.... \[0=-T^{2}+2T+8\]

  13. vacationeer
    • 3 years ago
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    of which can turnn into ..... \[0=(-T+4)(T+2)\]

  14. vacationeer
    • 3 years ago
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    so then T={4,-2} since you're dealing with only positive time (as time usually can only be positive)... you get T=4

  15. Garcia_123
    • 3 years ago
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    thanks!

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