Quote:
Originally Posted by 10isfreak
To give credence to your estimation, let's validate the approximation. Most pros hit at an angle less than 20 degrees and usually higher than 10 degrees (in terms of slope). We could deal with this using the tangent and a 100 units long side in a rectangular triangle.
Tan(20) = x/100 which leads us to 100 tan(20) = x = 36,4
Tan(10) = x/100 leads us to x =17,6
So, between 17,6 percent and 36,4 percent of the vector is oriented upward. With 35 m/s, it's 6,17 to 12,74 mm of upward movement.
We may have a good case to think that, indeed, the ball might move... An alternative possibility is that the ball moves along with the racket during this dwell time; yet an other possibility is that of a combination of sliding ball and moving ball. If we believe SpeedMaster's analysis (that much of spin is generated through the racket tilt and a contact bellow center  and by bellow, I mean closer to the ground, not the throat), only a fraction of this sliding effect would contribute to spin and this fraction should be less 50%, to say the very least.
A good way of finding if someone is educated or not is asking this person if things like this are complicated or not to understand. If he answers the later, he's either a genius who didn't reveal his conclusions or he sees the tip of the iceberg.

So I think what you saying is the ball is carried by the string some distance. In fact along the target arc, it might get carried more than two ball widths, and the upward part of that is probably what you are calculating. Along with that, the ball also slides along the strings a little.