Here's the question you clicked on:
Garcia_123
*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?
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.
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
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