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this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?

tried graphing it and it points downward which isn't right..

tried that too and it seems like it should point upward

There is no way it should be pointing upward with the angle at 60 deg

well this if for parametric equations so yes you can

not straight upward but to some degree yes it should be

my calculator has it pointing upwards

what did you use as the equations then?

Are you translating by 6 inches? That would be below the x axis pointed up.

but if you look at the table it appears to be positive for just a little bit

for the next part we should be solving\[h(t)=21.3t \sin(60)-16t^2-0.5=0\]to find the time in the air

of course we will get two answers and we want the larger one

Best wishes with this. My time's up for today. Bye guys.

later :)

but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?

what table?

hit 2nd and graph

i would get 1.3 seconds

sorry 1.13

thanks phi, I couldn't figure how to put parametric equations in wolfram.
"plot", huh?

oh wow yeah my graph looks nothing like that..dont know why

well i guess it sort of does match my answers, confused about the next part though

which part?

by r I mean the distance travelled

*should be
r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)

okay based on that i think it will make it to the receptacle...

double check it's 1.5 inches so 1.5/12 feet

right, because
\[21.3\cos(60)(1.125)=11.98\] which is in \[12\pm1.5\]

oh, I forgot conversions...

i did that at first too...

But it does hit the mark.

sorry, I'm going to bed, hopefully phi can help you. g'night!

its okay thanks!

and how exactly would you set that up?

plug into the first equation to find v

sorry but how do you solve for t when y = 0...im at 0 = (15/t cos (55)) t sin (55) - 16t^2 - 0.5

ok

so what would the answer for that be?