Question

(a) A small mail bag is released from a helicopter
that is descending steadily at 2.13 m/s.
After 4.38 s, what is the speed of the mail-
bag? The acceleration of gravity is 9.8 m/s2 .

(b) After the mailbag is dropped, the helicopter
continues descending for 1 s but then stops.
How far is the mailbag below the helicopter
at 4.38 s?

(c) What would be the speed of the mailbag if
the helicopter had been rising steadily at u =
2.13 m/s ? (Take down as positive.)
Answer in units of m/s

Answer 1

Part a)

Assume that downward is positive for this problem.

V=V_o + g*t

V_o = 2.13
g=9.8
t=4.38

V=2.13 + (9.8*4.38)
V = 45.1 m/s

part b)

x= V_o*t + (1/2)*g*t^2
x=(2.13*4.38)+(1/2)*9.81*4.38^2
x=103.43

note that the helicopter is going down at:
x1=2.13*4.38 = 9.33

X = x - x1
X = 103.43 - 9.33
X = 94.1 m

part c)
question states that assume down is positive, and that the helicopter is raising so:

V_o = -2.13

V=V_o + g*t
V= -2.13+(9.8*4.38)
V = 40.79 m/s