DAILY NEWS - A-Rod, who became the 29th player all-time to reach the 3,000-hit plateau on June 19 with a first-inning solo home run off Detroit's Justin Verlander, will be celebrated in a special pre-game ceremony on Sunday, Sept. 13, the Yankees announced Friday.
The team had previously downplayed Rodriguez’s accomplishments this season, declining to acknowledge his move up the all-time home run list when he tied Willie Mays for fourth place on the list on May 1. The Yankees were then embroiled in a dispute with A-Rod over a marketing agreement they had made with the slugger in 2007 centered around him moving up the list. That dispute has since been settled.
There's probably no greater testament to how this season has gone for A-Rod than this ceremony. Never in a trillion years would you have guessed that the Yankees would be throwing A-Rod a party to commemorate one of his milestones (I guess even though the bonus money situation for 661 home runs was sorted out it's still to murky a situation to market it). Yet here it is. And it makes plenty of sense. If you call yourself a Yankee fan then you've embraced A-Rod this year. He's been the key cog in the middle of this order and up until this recent slump has done nothing but help the team win. When he homered for his 3,000th hit this season the Yankee fan went ape shit, not just out of admiration for the achievement but in recognition of his total offensive renaissance this season.
The bottom line is that the ceremony makes sense. They get to reward A-Rod for being a good solider, a great teammate and a top-tier performer. They also get to sell tickets and t-shirts. Win-win for everyone.
PS - Coincidentally this is all going down on the first Sunday of the NFL schedule, same thing they did last year with Jeter. Toronto is in town that weekend. The Sunday game is at 1pm, same time as Jets-Browns. Should be a wild, wild day.
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