Question

The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 1.55 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

ive asked this question like 4 times..no one knows how to get it...no one

Answer 2

t = 1.55 --> h = 2.8(1.55)^3
this is mailbags initial height

initial velocity is zero since it is dropped

acceleration of gravity is -9.8 m/s^2
velocity = v(t) = -9.8t

position of bag -> y(t) = -4.9t^2 + h(initial height)

solve for t when y(t) = 0

Answer 3

in case you are wonderign how dumbcow did it, velocity measures the rate of change of position. So once you get the velocity function, you can integrate it to get the position function. It's an indefinite integral so there will be a constant, but that constant is just your original height.

Answer 4

i got the same answer i got last time....1.46 and it was wrong

Answer 5

thats y i wanted to check here lol

Answer 6

i get 1.4587, is it rounded correctly?
i will check my work

Answer 7

should i do 1.45?

Answer 8

no do they want 2 decimal places, if so it would be 1.46

Answer 9

it does say..all it says is what i pasted

Answer 10

doesnt**

Answer 11

well i double checked and i'm pretty sure my answer is correct
try using more decimals

Answer 12

said it was wrong and that was my last shot..im interested to see what the answer was..but thanks anyways =/

Answer 13

least we got the same answer and i wasnt going crazy

Answer 14

ok no problem, i guess i was wrong

Answer 15

me too lol

Answer 16

maybe its ...1.55+1.46 ??