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A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 3.00 s.

Required:
a. What is the speed of the mailbag?
b. How far is it below the helicopter?

Sagot :

Answer:

Explanation:

Initial velocity of mailbag u = 2 m/s

acceleration downwards a = g = 9.8 m/s²

time t = 3 s

a ) final velocity v = ?

v = u + at

= 2 + 9.8 x 3

= 31.4 m /s

b )

s = ut + 1/2 g t²

s is relative displacement of mailbag

u = relative initial velocity of mailbag = 0

relative acceleration = g = 9.8 m /s²

time t = 3 s

s = 0 + 1/2 x 9.8 x 3²

= 44.1 m

relative displacement of mailbag = 44.1 m .

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