Question

A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 5.00s what is the speed of the mailbag

Answer 1

The speed of the mailbag after 5.00s will be 2.00 m/s.

Step-by-step explanation:

The speed of the mailbag can be calculated using the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. In this case, the mailbag is released from a helicopter descending steadily at 2.00 m/s. Since the descent is steady, the acceleration is 0 m/s^2. Therefore, the final velocity of the mailbag after 5.00s will be 2.00 m/s.

Answer 2

The acceleration of gravity (on Earth) is 9.8 m/s² downward.

This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.

In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.

If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at

(2 m/s + 49 m/s) = 51 m/s .