In putting, the force with which a golfer strikes a ball is plannedso that the b

Question

In putting, the force with which a golfer strikes a ball is plannedso that the ball will stop within some small distance of the cup,say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie is more difficult than from adownhill lie. To see why, assume that on a particular greenthe ball decelerates constantly at 1.8 m/s2 goingdownhill, and constantly at 2.8 m/s2 going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculatethe allowable range of initial velocities we may impart to the ballso that it stops in the range 1.0m short to 1.0 m long of thecup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is moredifficult.

Explanation / Answer

then using teh formula vf2 -vi2 = 2ax here a = -2.8m/s2 vf = 0m/s,vi = vl, x = 6m then vl = sqrt[(2)(2.8m/s2)(6m)]    =5.8m/s
also let the velocity for the 1m long of the cup isvu. then using the formula vf2 -vi2 = 2ax here a = -2.8m/s2 vf = 0m/s,vu = vl, x = 8m then vu =sqrt[(2)(2.8m/s2)(8m)]    =6.7m/s therefore the range of the speed is 5.8m/s to6.7m/s
Now for the downhill lie 7m from the cup, let thevelocity for the 1m short of the cup is vl. then using the formula vf2 -vi2 = 2ax here a = -1.8m/s2 vf = 0m/s,vi = vl, x = 6m then vl =sqrt[(2)(1.8m/s2)(6m)]    =4.465m/s
also let the velocity for the 1m long of the cup isvu. then using teh formula vf2 -vi2 = 2ax here a = -1.8m/s2 vf = 0m/s,vu = vl, x = 8m then vu =sqrt[(2)(1.8m/s2)(8m)]    =5.37m/s therefore the range of the speed is 4.465m/s to5.37m/s
then using the formula vf2 -vi2 = 2ax here a = -2.8m/s2 vf = 0m/s,vu = vl, x = 8m then vu =sqrt[(2)(2.8m/s2)(8m)]    =6.7m/s therefore the range of the speed is 5.8m/s to6.7m/s
Now for the downhill lie 7m from the cup, let thevelocity for the 1m short of the cup is vl. then using the formula vf2 -vi2 = 2ax here a = -1.8m/s2 vf = 0m/s,vi = vl, x = 6m then vl =sqrt[(2)(1.8m/s2)(6m)]    =4.465m/s
also let the velocity for the 1m long of the cup isvu. then using teh formula vf2 -vi2 = 2ax here a = -1.8m/s2 vf = 0m/s,vu = vl, x = 8m then vu =sqrt[(2)(1.8m/s2)(8m)]    =5.37m/s therefore the range of the speed is 4.465m/s to5.37m/s
then using teh formula vf2 -vi2 = 2ax here a = -1.8m/s2 vf = 0m/s,vu = vl, x = 8m then vu =sqrt[(2)(1.8m/s2)(8m)]    =5.37m/s therefore the range of the speed is 4.465m/s to5.37m/s

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