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Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Up/Downhill corrections
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<blockquote data-quote="Jon A" data-source="post: 227489" data-attributes="member: 319"><p>This is incorrect. The above equation gives you the distance traveled, assuming an initial velocity of zero. 16.1 feet. It is provable by the following: If accelerating from zero at 32.2 f/s/s for 1 second obviously the terminal velocity is a * t = 32.2 * 1 = 32.2 ft/s. Since you started at zero and your acceleration was constant your average speed is (0+32.2)/2 = 16.1 ft/s. Average that velocity for 1 second and you travel 16.1 feet.</p><p></p><p>So the terminal velocity difference using a * t = 32.2*3.5=112.7 ft/s.</p></blockquote><p></p>
[QUOTE="Jon A, post: 227489, member: 319"] This is incorrect. The above equation gives you the distance traveled, assuming an initial velocity of zero. 16.1 feet. It is provable by the following: If accelerating from zero at 32.2 f/s/s for 1 second obviously the terminal velocity is a * t = 32.2 * 1 = 32.2 ft/s. Since you started at zero and your acceleration was constant your average speed is (0+32.2)/2 = 16.1 ft/s. Average that velocity for 1 second and you travel 16.1 feet. So the terminal velocity difference using a * t = 32.2*3.5=112.7 ft/s. [/QUOTE]
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Up/Downhill corrections
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