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I'm not sure what to make of the rest of your post.
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Two bullets shot at the same muzzle velocity, but with one subjected to gravity and one not, would both reach a point along the x-axis at the same time. The only difference is that the one not subjected to gravity would be at a higher point along the y-axis. This is despite the fact that the gravitationally affected bullet would have traveled a greater distance due to its more curved path. The important concept to grasp is that the motion made in two directions by this bullet takes place simultaneously, so that it doesn't take any more time to cover the greater distance. Thus I and the laws of physics disagree with Dexter's premise that the ball which traveled a greater distance would have had to move at a higher average speed in order to reach the destination at the same time. |
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"To win the game is great. To play the game is greater. But to love the game is the greatest of all." |
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The old movie "Take Me Out To The Ballgame" was on Turner Classic the other day. It stars Frank Sinatra and Gene Kelly. In the movie, they were part of a double play combination, O'Brien to Ryan to Goldberg. Gotta love it.
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Yom HaShoah |
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Good Lord! I'm not writing what I'm thinking today. I've got to be too tired from staying up for the fireworks last night. Here's what that sentence should have said:
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BTW while I took that class in DC, it was taught by the visiting Dean of Yale. |
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To respond to your post: Let's take two balls. (A) is a "line drive" - hit parallel to the ground, and we disregard the effect of gravity. (B) is a more traditional HR ball - hit in a parabolic arc (where we do have gravity). We'll assume that the horizontal component of both of their speeds are equal (let's say 10 m/s, for example). If this is the case, B will have to have a greater overall AVERAGE speed, and we can see this in two ways. First - geometrically, by combining vectors. Both A and B have the same horizontal speeds. However, ball A also has a vertical component to its speed. When we add the vectors together, the sum of the speeds of A will be greater than the speed of B. Second - algebraically. In Mick's original example, he simplified the HR ball (B) to a line-drive style hit (A). Given this, the time for A and B to travel the same horizontal distance is the same. However, ball A travels a further overall path than B does. If two objects travel different distances in the same amount of time, they must have different speeds. (With the ball that travels further obviously having the faster speed).
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"To win the game is great. To play the game is greater. But to love the game is the greatest of all." |
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"To win the game is great. To play the game is greater. But to love the game is the greatest of all." |
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[2 out, bottom of 5th, 1-2 on Marcus Thames --> Jack City. 1-8 and waiting for a new Red Sox pitcher] Last edited by mick; Fri Jul 06, 2007 at 07:41pm. |
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Oh, well. We're still the only AL East team above .500.
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"To win the game is great. To play the game is greater. But to love the game is the greatest of all." |
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Did anyone see this crazy game? 34 runs were scored and it wasn't played in Colorado or Cinci! PS Dexter, those are the same thoughts that I was having, but I'm not reaching the same conclusions as you. Essentially, I'm thinking that the g vector will cancel itself out from the half of the parabola that is ascending and the half that is descending, leaving one with the same average. Let me ponder it for a bit and get back to you. |
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Where's Billy Pierce, Early Wynn, Dick Donovan and Wilbur Wood when you need them?
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It's 11:30, and I'm still watching the bottom of the 13th. (Papelbon isn't helping things along any by checking on Sheffield at 1B continuously.) Of course - as soon as I type that, Shef steals second. Could be worse - apparently Mets-Astros is in the bottom of the 15th.
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"To win the game is great. To play the game is greater. But to love the game is the greatest of all." |
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