Question
A brick is vertically dropped from a building at 2m/s to reach the ground after 4s. Calculate the distance the brick fell
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GPT 3.5
To calculate the distance the brick fell, we can use the formula for distance traveled under constant acceleration:
distance = initial velocity * time + 0.5 * acceleration * time^2
In this case, the initial velocity is 2 m/s, the time is 4 s, and we need to find the acceleration.
Since the brick is dropped vertically, we assume it falls under gravitational acceleration, which is approximately 9.8 m/s^2.
distance = 2 m/s * 4 s + 0.5 * 9.8 m/s^2 * (4 s)^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 16 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance ≈ 8 m + 499.2 m
distance ≈ 507.2 m
Therefore, the distance the brick fell is approximately 507.2 meters.
distance = initial velocity * time + 0.5 * acceleration * time^2
In this case, the initial velocity is 2 m/s, the time is 4 s, and we need to find the acceleration.
Since the brick is dropped vertically, we assume it falls under gravitational acceleration, which is approximately 9.8 m/s^2.
distance = 2 m/s * 4 s + 0.5 * 9.8 m/s^2 * (4 s)^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 16 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance = 8 m + 0.5 * 9.8 m/s^2 * 256 s^2
distance ≈ 8 m + 499.2 m
distance ≈ 507.2 m
Therefore, the distance the brick fell is approximately 507.2 meters.
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