Question

The height of a helicopter above the ground is given by h = 2.85t3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

wow

Answer 2

i mean i am looking for the answer

Answer 3

later

Answer 4

\[a=10m s ^{-2}\]\[u=0 \]
s=2.85m
\[v^2-u^2=2as\]
\[v^2= 20m/s^2 * 2.85\]
\[v^2= 57m^2 s ^{-2}\]
\[\sqrt{57} = 7.55\]
\[a=v/t [u=o]\]
\[10m/s^2 = 7.55m/s \div t\]\[t= 5.77/10 = 0.57s\]
the time it reaches the ground from releasing from the sky = 0.57s
after the heli take off is 1.65+0.57s
=2.22s