Posted by Hunter on .
A small mailbag is released from a helicopter that is rising steadily at 2 m/sec. a) how long does it take the mailbag to stop moving? b) how high will the mailbag have risen in the time found in part a? c) how far did the helicopter move in the time found in part a?
The mailbag moves upwards with the initial velocity v0=2 m/s.
Its motion is decelerated
h = v0t - (gt^2)/2
v = v0 - gt.
If v=0, v0=gt and t = v0/g = 2/9.8=0.2 s.
h = 2•0.2-(9.8•0.04)/2 = 0.204 m
Helicopter moves uniformly, therefore,
H = vt = 2•0.2 = 0.4 m.