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A small mailbag is released from a helicopter that is descending steadily at 1.37 m/s. After 5.00 s, what is the speed of the mailbag?

Answer :

Final answer:

The speed of the mailbag after 5 seconds is 47.63 m/s, downwards. This is calculated using the physics equation for an object in free fall and taking into consideration that the helicopter is already moving downwards.

Explanation:

In Physics, the speed of an object in free fall, ignoring air resistance, is determined by the equation v = u + gt, where 'v' is the final velocity, 'u' is the initial velocity, 'g' is acceleration due to gravity (approximately 9.8 m/s² on earth), and 't' is the time. In this case, the initial velocity 'u' is -1.37 m/s (negative because the velocity is directed downwards), 'g' is 9.8 m/s², and 't' is given as 5 seconds.

Applying these values to our equation gives v = -1.37 m/s + (9.8 m/s² * 5 s) = 47.63 m/s (downwards). This is the speed of the mailbag after 5 seconds.

Learn more about Final Velocity of Falling Object here:

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