# Physics

(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
2.13 m/s ? (Take down as positive.)

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2. 👎
3. 👁
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

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2. 👎

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