A golfer needs your help to choose which club to use for their next stroke. Some things you need to know include:

Consider that the ball will travel directly between where it starts and the hole.

The ball’s speed can be separated into a vertical component and a horizontal component. When the actual speed is represented by the hypotenuse of a right triangle, the legs of the triangle represent those components. (Recall the Pythagorean Theorem and similar triangles from Geometry.)

Irons (a type of golf club) are numbered 1-9.

The face of each club is angled differently. The greater the number of the club, the more it directs the ball into the air. As an approximation, consider that the ratio of vertical to horizontal movement is the club number : 6.

The ball will bounce only once, but not as high. When it bounces, it will only reach 1/4 of the height it reached before the bounce because its speed will be cut in half.

After the bounce, the ball will essentially roll a distance equal to the product of…

the reciprocal of the lift ratio, and

the horizontal distance gained during the bounce.

The height reached by the ball, in yards, can be modeled by the equation:

[tex]h(t) = -\frac{16}{3}t^{2} + v_{0} t[/tex]

where v0 is the initial VERTICAL velocity (yards/second)

and t is time measured in seconds.

The ball starts 150 yards from the hole and

the golfer can deliver enough force to produce a maximum speed of 40 yds/sec.

Analyze at least two different clubs to:

determine the appropriate percentage of the maximum force that should be applied by the golfer to get the ball to the hole, and

find the maximum height reached by the ball in each scenario.

Given a golf club number 'n' hits the ball with a vertical : horizontal velocity ratio of n, and a golfer can apply a maximum force sufficient to hit a ball with a velocity of 40 yd/s, you want the percentage of maximum force required to hit a ball 150 yards using two different golf clubs in the range n=1–9. The ball bounces to 1/2 its initial velocity one time, then rolls a distance 1/n times the distance traveled on the bounce. The equation for ballistic motion applies when the ball is in the air.

Hang time

For initial vertical velocity v0, the height equation is ...

h(t) = -16/3t² +v0t

Then the time required for the ball to return to zero height is ...

16/3t² = v0t

t = 3·v0/16

Distance to bounce

The horizontal distance to the bounce is

distance = speed · time

d1 = (v0·6/n)·(3·v0/16) = 9·v0²/(8n)

where v0·6/n is the horizontal speed for club number 'n'.

Bounce distance

We see the distance traveled is proportional to the square of the vertical velocity. After the bounce, the ball has 1/2 its initial vertical velocity, so its hang time is 1/2 what it was. If we assume the horizontal velocity is unaffected by the bounce, then the distance gained during the bounce is ...

d2 = d1/2 = 9·v0²/(16n)

Roll distance

The distance the ball rolls is 1/n times the bounce distance:

d3 = (1/n)d2 = 9·v0²/(16n²)

Total distance

The total distance the ball travels is ...

d = d1 +d2 +d3

d = 9·v0²/(8n) +9·v0²/(16n) +9·v0²/(16n²)

d = (9·v0²/(16n²))·(3n +1)

Force

We want to know the fraction of full force required to achieve a distance of 150 yards.

Solving the distance equation for v0², we have ...

150 = (9·v0²/(16n²))·(3n +1)

v0² = 150(16n²)/(9(3n+1)) = 800n²/(3(3n+1))

The square of the total ball velocity is the square of the maximum velocity multiplied by the fraction of maximum force. (Energy is proportional to force and to the square of velocity.) The Pythagorean theorem tells us the square of the total velocity is the sum of the squares of the horizontal and vertical velocities:

Vtot² = v0² +(v0·6/n)² = v0²·(1 +(6/n)²)

Vtot² = k·Vmax² = v0²·(1 +(1/n)²)

Using 40 yd/s as Vmax, we find v0² to be ...

k·(40²) = v0²·(1 +(6/n)²)

v0² = 1600k·n²/(36 +n²)

Using this expression in the above equation for v0², we can solve for k in terms of n

1600k(n²)/(36+n²) = 1600n²/(6(3n+1))

k = (n² +36)/(6(3n +1))

For the different golf club numbers, the force fractions will be ...

club 2: (2² +36)/(6(3·2 +1)) = 40/42 = 20/21 ≈ 95.2%
club 8: (8² +36)/(6·(3·8 +1)) = 100/150 = 2/3 ≈ 66.7%

(Golf club number 1 will not get the ball to the hole with the force available.

Height

The height an object will reach is the height at which its potential energy is equal to its initial vertical kinetic energy. That gives a height of ...

h = v0²/(2g) . . . . . where g = 32/3 yd/s²

h = (3/64)v0² = (3/64)1600k·n²/(36 +n²) = 75kn²/(n²+36)

Assuming k is the value necessary for the ball to reach the hole, this gives ...

h = (75n²/(n²+36))·(n²+36)/(6(3n+1)) = 12.5n²/(3n +1)

Then for clubs 2 and 8, the maximum height is ...

club 2: h = 12.5(2²)/(3·2 +1) = 50/7 = 7 1/7 yards
club 8: h = 12.5(8²)/(3·8 +1) = 800/25 = 32 yards

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Additional comment

The attached graph shows the ball path to the end of the first bounce for clubs 2 and 8.

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