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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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