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*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?

Mathematics
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where are the "T's" in the equation?
cuz right now with what you have... it would just be.... 476 no matter what... which is impossible
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.

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Other answers:

O.o.......i mean the equation would work better if it was lik \[h(t)=-16(T^{2})+32T+188\]
find time at max height
take the first dervivative
that would make more sense... in which you would substitute h(t) for 60 and then solve from there
@mandonut ..... its not a max question... its asking for the position of the ball
*coin
so then it would be \[60=-16(T^{2})+32T+188\]
pull the 60 over to get.... \[0=-16(T^{2}) +32T +128\]
now you can take out 16 from the whole equation to get.... \[0=-T^{2}+2T+8\]
of which can turnn into ..... \[0=(-T+4)(T+2)\]
so then T={4,-2} since you're dealing with only positive time (as time usually can only be positive)... you get T=4
thanks!

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