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A bag is dropped from a hovering helicopter. When the bag has fallen -13 m, what is its velocity?

User Xashru
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Final answer:

The velocity of a bag that has fallen -13 meters from a hovering helicopter is found using the kinematic equation. Assuming a starting velocity of 0 m/s and an acceleration due to gravity of -9.80 m/s², the final velocity is calculated to be -16 m/s, indicating a downward motion.

Step-by-step explanation:

The question is asking to determine the velocity of a bag that has been dropped from a hovering helicopter and has fallen a distance of -13 meters. To solve this, we will use the kinematic equation v² = v₀² + 2a(y - y₀), where v is the final velocity, v₀ is the initial velocity, a is the acceleration due to gravity (which is -9.80 m/s² since it's acting downwards), y is the final position, and y₀ is the initial position. In this case, we will take the initial velocity (v₀) to be 0 m/s since the bag is dropped and not thrown, and the initial position (y₀) to be 0 m since the bag starts from rest at the point of release.

Using the kinematic equation with the known values:

v² = (0 m/s)² + 2(-9.80 m/s²)(-13 m) = 0 + 255.2 m²/s²

Taking the square root of both sides, we find that:

v = ±16 m/s

As the bag is falling downwards, we will take the negative root, which indicates the direction of the velocity is also downwards. Hence the velocity of the bag after falling -13 meters is -16 m/s.

User Salt
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