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A bag is dropped for a hovering helicopter. When the bag has fallen for 2.0 s, what is the bag's velocity?

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Final answer:

The bag's velocity after falling for 2.0 seconds from a hovering helicopter is 19.6 m/s when calculated with the kinematic equation for objects in free fall: v = at, with acceleration due to gravity (9.8 m/s²).

Step-by-step explanation:

The question is asking for the velocity of a bag 2.0 seconds after it is dropped from a hovering helicopter. To find the velocity, we can use one of the kinematic equations that relates velocity, acceleration, and time:

v = vo + at

Where v is the final velocity, vo is the initial velocity, a is the acceleration (9.8 m/s2 for objects in free fall due to gravity), and t is the time. Since the initial velocity (vo) of the bag is zero (as it is dropped and not thrown), the equation simplifies to:

v = at

Thus, the bag's velocity after 2.0 seconds is:

v = (9.8 m/s2) × (2.0 s) = 19.6 m/s

The bag will have a velocity of 19.6 m/s after falling for 2.0 seconds, assuming that air resistance is negligible.

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