A falling body will accelerate at a constant rate of speed, just because he specified the weight of the falling body doesn't mean that it's wrong, it just means that the fall of a 150 pound man will be be the same.
to smileygurrl: I guess it depends on how you read it. That's the way I read it, you may read it differently. It seems to me that he made the weight of the body a factor.
All right well the gravitational acceleration depends on what planet you are on. The acceleration on earth depends on the distance from the earth's center of gravity and the Earth's mass in kilograms, just as in any planet, and Earth pulls objects toward her at a rate of 9.81 m/(seconds)^2.
Having that said, the equation for this sort of thing is:
d = V(initial) * t + 1/2a*t^2
Note that in this equation, the mass is not a factor. That is because a man weighing 150 pounds would present little air resistance for it to change anything. Imagine a heavy steel ball that weighs the same falling alongside the man. Though I admit this equation is wrong for great distances: an object falling would eventually reach terminal velocity, where the Force gravity down equals the Force air pushing up. When two forces are equal, there is no acceleration, only a constant velocity. The velocity would increase less and less until it reaches that terminal velocity. They would hit at times so close that we would say that they fell at the same time. Now, we have to convert pounds into kilograms and feet into meters, so :
150 lbs = 68.04 kilograms
300 feet = 91.44 meters (almost a kilometer...ouch)
Now, using a as 9.81, we get the equation
91.44 = 0 * t + 1/2 * 9.81 * t^2
Initial velocity is 0, as he is dropped. Imagine him standing on something at a height of 300 feet, and that thing suddenly disappears. Using the equation above, t ( the positive value) comes out to be 4.32 seconds, which is <5.
Figuring air resistance would get you an answer that would not change much. I know that it won't be greater than 5.
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