A rock is thrown from the top of a 48 foot tall building. The distance, d, in feet, between the rock and the ground t seconds after the rock is thrown is given by the equation d=-16t^2+32t+48. a)How far off the ground is the rock 2 seconds after it was thrown? b)How long after the rock is thrown does the rock hit the ground?

plug in 2 for t to find distance plug in 48 for d to find time

okay great! thanks a bunch!!

Do I then foil the problem?

48=-16t^2+32t+48 is this how you would write it to solve it???

yea umm u can take out an 16 so ud get -t^2+2t+3

so then the problem will look like this 48=-16(t^2+2t+3)???

how far did it travel when u solved for t=2?

I got 48 ft

yea thats wat i thought.. so if its a 48ft tall building.. and after 2 seconds it goes 48 ft... .... u c where this is going?

ummmm.......so it will be 3 sec??

lol it went 48 ft and the building is only 48 ft tall so it must have it the ground at 2 seconds right? unless this was a phyiscs question then thers a horizontal toss equation

well I don't know how to do the second equation right them lol

hm no i think 2 would be the answer then cuz if it was physics then the other equation is x=vt and u dont have a v in ur original question

so do i write the question as 48=-16t^2+32t+48 to find d?? I then pull -16 out and minus 48 from both sides??

o.. nvn u were right earlier the answer is 3 because u plug in 0 for distance in d and factor to get the roots of 0 and 3 and since u cant have 0 as ur answer the answer is 3 i didnt read ur original right cuz i was distracted sry :P

Great! thanks for your help!!

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