In putting, the force with which a golfer strikes a ball is planned so that the
ID: 1411053 • Letter: I
Question
In putting, the force with which a golfer strikes a ball is planned so 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 (that is, putting the ball downhill, see (Figure 1) ) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at 1.9 m/s2 going downhill, and constantly at 2.4 m/s2going uphill.
Figure 1 of 1
Part A
Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup.
Express your answers using two significant figures separated by a comma.
Part B
Suppose we have an downhill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup.
Express your answers using two significant figures separated by a comma.
Explanation / Answer
Part A
v2 = u2 + 2 a s
v = 0 , s = 8 m or 6 m
downhill acceleration a = -1.9 m/s2
u = sqrt (-2as)
u = sqrt (2 x 1.9 x 8)
u = 5.51 m/ s
or
u = sqrt (2 x 1.9 x 6)
u = 4.77 m/s
The range of initial velocities are 4.8 m/s to 5.5m/s
Part B
distance = 7 +1 or 7-1 = 8m or 6 m
Uphill acceleration = - 2.4 m /s2
Final velocity v = 0
Using Newton's laws of motion
v2 = u2 + 2 as
u = sqrt (-2 as)
u = sqrt (2 x 8 x 2.4)
u = 6.197 m/s
or
u = sqrt (2 x 6 x 2.4)
u = 5.366 m/s
Thus initial velocities are in range of 5.4 m/s to 6.2 m/s