Question

A flaming arrow is shot into the air to mark the beginning of a festival. its height, h, in meters, after t seconds can be modeled by the function h(t) = -4.9t^2 +24.5t +2
d) how long does it take the arrow to return to the ground
the textbook's answer is 5.08 s

Answer 1

what you're given is the position equation

for time t, you get the height of the arrow

so the height of the ground is zero

-4.9t^3 +24.5t +2 = 0

solve for t

also, i believe the equation is wrong

it should be t^2 instead of t^3

Answer 2

lol yes that is a typo i fixed it...i tried to solve for t but the furthest i get is \[-2 = -4.9t^{2} + 24.5t\] and i'm not sure what to do after this

Answer 3

am i supposed to divide out the -4.9 or 24.5? and even if i do, how do i get rid of the extra t?

Answer 4

you dont move the -2 to the other side

since its a quadratic equation, you can either factor which will be hard
or use the quadratic equation to solve for t (which i recommend)

Answer 5

alright i'll try the quadratic quadratic eqtn

Answer 6

omg it worked!! thank youuuu

Answer 7

now you know how to solve these problems
dont forget it

Answer 8

i mean the process really isnt that difficult

Answer 9

true, i guess its just hard to remember when to use what formulas

Answer 10

if you take calculus, problems like these become waaaaaay easier

you can solve them in seconds :)

Answer 11

lol this is from calculus...got my exam in a couple of days :P