For a ball dropped from rest with g = 9.8 m/s^2, what is the distance fallen after t = 3 s?

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Multiple Choice

For a ball dropped from rest with g = 9.8 m/s^2, what is the distance fallen after t = 3 s?

Explanation:
When an object falls under constant downward acceleration from rest, the distance it travels is s = ut + (1/2) a t^2. Here, the initial speed u is 0, and the acceleration a is g = 9.8 m/s^2, so s = (1/2) g t^2. Plugging in t = 3 s gives s = 0.5 × 9.8 × 9 = 44.1 m. So the ball falls 44.1 meters. The value 29.4 m/s would be the speed after 3 seconds (v = gt), which is a velocity, not a distance. The other numbers don’t fit the distance-from-rest formula.

When an object falls under constant downward acceleration from rest, the distance it travels is s = ut + (1/2) a t^2. Here, the initial speed u is 0, and the acceleration a is g = 9.8 m/s^2, so s = (1/2) g t^2.

Plugging in t = 3 s gives s = 0.5 × 9.8 × 9 = 44.1 m. So the ball falls 44.1 meters.

The value 29.4 m/s would be the speed after 3 seconds (v = gt), which is a velocity, not a distance. The other numbers don’t fit the distance-from-rest formula.

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