Puzzle
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Ponder This Challenge - March 2001 - Golf - hit in one

Ponder This Challenge:

Suggested by Mike Cargiulo

We're playing golf in the Tyrolean Alps.
The air is so thin up here that air resistance is negligible.
And the course is laid out a bit strangely...

The first hole is 150 yards, due horizontal.
(The cup is on the same vertical level as the tee.)
We hit the ball at top speed, at the correct angle to maximize distance, and find that it precisely reaches the cup. Hole in one.

The second hole is also 150 yards, but due vertical: the cup is directly below the tee.
This time we require no force at all: we can merely drop the ball into the cup.

The third hole is a bit tricker: 150 yards, at a 45 degree angle below the horizon. At what angle should we hit the ball, and at what speed (as a percentage of the speed required on the first hole), if we wish to minimize the speed required?

David Buller (age 11) has a clever solution to the problem as stated:

Simply give the ball a very slight nudge (so the speed is essentially zero).  Gravity will take the ball rolling down the hill for a hole-in-one.  So the answer is speed=0 and angle=-45 degrees.

The rest of you, let's restrict ourselves to a ball exactly hitting the cup at the end of its parabolic arc, without rolling. Thanks, David.

We will post the names of those who submit a correct, original solution! If you don't want your name posted then please include such a statement in your submission!

We invite visitors to our website to submit an elegant solution. Send your submission to the ponder@il.ibm.com.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solution

  • On the first hole, suppose the ball's initial speed is (v), at an initial angle (b) up from the horizontal. Its horizontal velocity is then (v cos b). Its initial vertical velocity is (v sin b). Letting (g) be the acceleration due to gravity, we see that after time (t = 2 v sin b / g) the vertical velocity will have changed from (v sin b) to (- v sin b), and the ball will have returned to its initial vertical position. Let (D=150 yards) denote the horizontal distance of this hole, and observe that in time (t) the ball has travelled a horizontal distance of (t v cos b), leading to the equation:

    D = (v^2 /g) 2 sin b cos b,
    or
    D = (v^2 /g) sin(2b).

    To maximize the horizontal distance for a given initial speed (v), we must maximize (sin(2b)), which we do by setting (b=45 degrees), whence (sin(2b)=1), and

    v = sqrt ( D g ).

    On the third hole, denote the initial speed as V, the initial angle above the horizon as B, and the total time as T.
    The horizontal distance will be

    D / sqrt(2) = T V cos B.

    The average vertical velocity will be

    V sin B - (T/2) g,

    so that the total vertical ascent will be

    • D / sqrt(2) = T ( V sin B - (T/2) g ).

    Solving for (V):

    V^2 = D g / ( 2 sqrt(2) * (cos B) * (sin B + cos B) ).

    Use the identity:

    (cos B)*(sin B + cos B) = (1/2) + (1/sqrt(2))*sin(2B+45)

    to deduce that (V) will be minimized when (B=22.5 degrees), with a ratio

    V/v = sqrt ( 1 - 1/sqrt(2) ) = 0.5412.

    On the third hole, the golfer hits the ball at 54.12% of the speed that was required on the first hole.

    If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solvers

  • Henry Bottomley (03.01.2001 @ 10:57 AM EDT)
  • Aron Fay (03.01.2001 @ 11:18 AM EDT)
  • Derek Kisman (03.01.2001 @ 1:45 PM EDT)
  • Marian Olteanu (03.01.2001 @ 2:04 PM EDT)
  • John Hamilton (03.01.2001 @ 2:18 PM EDT)
  • Colin Bown (03.01.2001 @ 2:24 PM EDT)
  • Sorin Ostafiev (03.01.2001 @ 8:10 PM EDT)
  • Dharmashankar Subramanian (03.01.2001 @ 10:39 PM EDT)
  • Ionel Santa (03.02.2001 @ 5:58 AM EDT)
  • Francis Golding (03.02.2001 @ 6:25 AM EDT)
  • Ionut Breaz (03.02.2001 @ 8:06 AM EDT)
  • Michael Malak (03.02.2001 @ 10:14 AM EDT)
  • Daniel Bitin (03.02.2001 @ 12:28 PM EDT)
  • Andrew Froggatt (03.03.2001 @ 12:11 PM EDT)
  • Ming Song (03.03.2001 @ 3:03 PM EDT)
  • Dave Biggar (03.04.2001 @ 8:22 AM EDT)
  • Prithu (03.04.2001 @ 12:10 AM EDT)
  • Ariel Landau (03.04.2001 @ 8:30 AM EDT)
  • Dieter Verhofstadt (03.05.2001 @ 5:12 AM EDT)
  • Vince Lynch (03.05.2001 @ 3:51 PM EDT)
  • Xuan Baldauf (03.05.2001 @ 9:08 PM EDT)
  • Stephane Higueret (03.06.2001 @ 3:49 AM EDT)
  • Christie Bolton (03.06.2001 @ 3:50 PM EDT)
  • Krishna Kumar (03.07.2001 @ 10:56 AM EDT)
  • Karthikeyan Bhasyam (03.07.2001 @ 12:23 PM EDT)
  • Prabhat Agarwal (03.07.2001 @ 1:28 PM EDT)
  • Darryl Buller (03.07.2001 @ 4:29 PM EDT)
  • Jonathan Wildstrom (03.07.2001 @ 6:55 PM EDT)
  • C A Subramanian (03.07.2001 @ 10:36 PM EDT)
  • Douglas Schrag (03.08.2001 @ 10:20 AM EDT)
  • Shmuel Spiegel (03.08.2001 @ 12:08 PM EDT)
  • Vic Monell (03.08.2001 @ 3:50 PM EDT)
  • Alexey Vorobyov (03.09.2001 @ 3:54 AM EDT)
  • Nicusor Timneanu (03.09.2001 @ 9:57 AM EDT)
  • Joshua Zirbel (03.09.2001 @ 3:59 PM EDT)
  • Mihai Cioc (03.11.2001 @ 2:10 PM EDT)
  • Paul Thomas (03.11.2001 @ 9:27 PM EDT)
  • Ralf Seliger (03.14.2001 @ 1:34 PM EDT)
  • Scott Ruland (03.19.2001 @ 9:43 AM EDT)
  • Marco Aiello (03.21.2001 @ 7:22 PM EDT)
  • Ross Millikan (03.22.2001 @ 5:16 PM EDT)
  • Jesus Sanz (03.23.2001 @ 9:18 AM EDT)
  • Paul Vandenkeybus (03.24.2001 @ 8:41 AM EDT)
  • Brian Bitto (03.27.2001 @ 3:48 AM EDT)
  • Steve Walter (03.27.2001 @ 10:49 AM EDT)

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