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To calculate the velocity of an object just before it touches the ground when dropped from a height of 80 meters, we can use the principles of motion and the equations of motion under constant acceleration. Assuming the object is subject only to the acceleration due to gravity, we can use the following equation:

v^2 = u^2 + 2as

where: v = final velocity (which we need to find) u = initial velocity (which is 0 when the object is dropped) a = acceleration due to gravity (approximately 9.8 m/s^2) s = distance traveled (which is the height of the object, 80 meters)

Plugging in the known values, we have:

v^2 = 0 + 2 * 9.8 * 80

v^2 = 1568

Taking the square root of both sides, we find:

v ≈ 39.6 m/s

Therefore, just before touching the ground, the object will have a velocity of approximately 39.6 meters per second.

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