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A pre-1982 penny with a mass of 3.1 grams is dropped from a skyscraper of 1250 feet. Find the object's acceleration after 5 seconds. If a post-1982 penny with a mass of 2.5 grams is dropped at this point, how far does it travel before the first penny hits the ground?
The first answer is 9.8 m/s2 because the acceleration is only due to gravity and is therefore independent of time - it is always 9.8 m/s2 thoroughout the flight of the penny. The second question however - seems straightforward, but I don't know how they got the answer that they did. I figured that it would first be necessary to calculate the total time in flight (which is independent of mass b/c they both start at zero velocity and are equally accelerated by gravity). In order to do this, you must first convert the 1250 feet to meters - there are 3.3 feet per meter - therefore 1250 feet is about 378 meters. Then, using x=Vot + 1/2at^2, you get approximately 8.5 seconds for the time from the drop to hitting the ground. If the 2nd penny is dropped 5 seconds after the first one - then there is 3.5 seconds for it to fall before the first penny hits the ground. Using the same equation with 3.5 seconds for t and solving for x, I get about 61.25 meters or about 202 feet.
The answer in the back is 1250 feet. I do not understand how they arrived at that answer unless they are saying 'at this point' means 'at the same height as the first penny' instead of 'after the first penny falls 5 seconds.' Any other ideas?
The first answer is 9.8 m/s2 because the acceleration is only due to gravity and is therefore independent of time - it is always 9.8 m/s2 thoroughout the flight of the penny. The second question however - seems straightforward, but I don't know how they got the answer that they did. I figured that it would first be necessary to calculate the total time in flight (which is independent of mass b/c they both start at zero velocity and are equally accelerated by gravity). In order to do this, you must first convert the 1250 feet to meters - there are 3.3 feet per meter - therefore 1250 feet is about 378 meters. Then, using x=Vot + 1/2at^2, you get approximately 8.5 seconds for the time from the drop to hitting the ground. If the 2nd penny is dropped 5 seconds after the first one - then there is 3.5 seconds for it to fall before the first penny hits the ground. Using the same equation with 3.5 seconds for t and solving for x, I get about 61.25 meters or about 202 feet.
The answer in the back is 1250 feet. I do not understand how they arrived at that answer unless they are saying 'at this point' means 'at the same height as the first penny' instead of 'after the first penny falls 5 seconds.' Any other ideas?
