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A small mailbag is released from a helicopter that is descending steadily at 1.95 m/s.

(a) After 3.00 s, what is the speed of the mailbag?

(b) How far is it below the helicopter?


Sagot :

Answer:

(a) 31.38 m/s

(b) 38.3 m

Explanation:

(a)

We know that acceleration due to gravity, g, is 9.81 m/[tex]s^2[/tex].

This means that the descending speed of the mailbag will increase by 9.81m/s every second.

So after 3 seconds its speed will have increased by,

3 x 9.81 m/s = 29.43 m/s

We also already know that the mailbag was descending at a speed of 1.95 m/s before being released from the helicopter.

We can now find the final velocity of the mailbag by adding the change in velocity (29.43 m/s) to the initial velocity (1.95 m/s),

1.95m/s + 29.43 m/s = 31.38 m/s

(b)

We will use the equation [tex]s = 1/2gt^2[/tex], where

s is the distance travelled,

g is the accelaration due to gravity and

t is the time taken to travel the distance.

Using this equation we can determine that in 3 seconds the mailbag has fallen,

[tex]1/2 *(9.81 m/s^2) *(3 s)^2 = 44.145 m[/tex]

In these 3 seconds the helicopter has descended,

3 s x 1.95 m/s = 5.85 m

So the mailbag is now

44.145 m - 5.85 m = 38.3 m (3 s.f.)

below the helicopter.

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