On 4/10/07, David VanHorn wrote: > Actually, this is EE, but I can't explain why. > > Does anyone know where to find the acceleration of a baseball when it is hit? > I'm looking for high end numbers like in a major league home run. > Gee and Jerk if I can get it. Order of magnitude calculation: Incoming baseball velocity: -90mph (good fastball) (note negative sign) Homerun distance: 350 feet Vy(t) = V sin(theta) - gt Vx(t) = V cos(theta) At end of flight Vy(t) = -Vy(0) V sin(theta) - gt = -V sin(theta) 2 * V sin(theta) = gt t = 2 * V sin(theta) / g x distance: x(t) = Vx(t) * t = V cos(theta) * 2 * V sin(theta) / g Assume no air resitance, optimum angle on trajectory (45 degrees) x(t) = 350 = 2 (V^2) sin(theta) cos(theta) / g V = sqrt(g * d / (2 * cos(theta) * sin(theta))) V ~= 106 ft/second Assume impact takes place for a duration of 1/10th of a second: a = (V1 - V0) / t a = 106 - (-90) / 0.10 a = 1960ft/second^2 a ~= 61.25 g Note that this is an AVERAGE acceleration, the real acceleration will not be uniform over time, and there will be an instantaneous acceleration much larger than this average. Bill -- Psst... Hey, you... Buddy... Want a kitten? straycatblues.petfinder.org -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist