Extreme Putting

Classical Mechanics Level 5

A world-class golfer wants to show off his extreme putting skills on a non-standard putting green in the shape of the following plane (distances in meters):

$\large{-x - y + 3z = 0}$

The golfer is standing at $(x,y,z) = (0,0,0)$ (with the golf ball at the same location), and the hole is positioned at $(x,y,z) = (10,20,10)$ .

Gravity is $10 \, \text{m/s}^2$ in the $-z$ direction.

If the ball enters the hole $10$ seconds after being hit by the golfer, what is the ball's initial speed (in $\text{m/s}$ )?

Details and Assumptions:
- The ball can be modeled as a massive point-particle
- The ball stays on the green the entire time, and moves without losses
- The ball's initial velocity is parallel to the green
- In order to enter the hole, the ball must simply intersect the hole's coordinates

The answer is 23.6758.

1 solution

Steven Chase
Jan 26, 2018

Here is a high-level outline of the solution. The decompositions described are performed using linear algebra.

Determine normal and tangential unit vectors to the plane (all orthogonal to each other) :

$\text{Normal:} \, \vec{N} \\ \text{First Tangential:} \, \vec{t_1} \\ \text{Second Tangential:} \, \vec{t_2}$

Represent the gravity as a superposition of these vectors (solve for component accelerations)

$\vec{g} = a_N \, \vec{N} + a_1 \, \vec{t_1} + a_2 \, \vec{t_2}$

Planar acceleration (this is now effectively a 2D problem)

$\vec{a} = a_1 \, \vec{t_1} + a_2 \, \vec{t_2}$

Represent the hole (xf,yf,zf) coordinates in terms of the new system coordinates (with axes aligned with the tangent vectors)

$(x_f,y_f,z_f) \rightarrow (c_1,c_2)$

Kinematic Equations (final time $t_f$ is known; $v_1$ and $v_2$ are initial component velocities):

$c_1 = v_1 \, t_f + \frac{1}{2} a_1 \, t_f^2 \\ c_2 = v_2 \, t_f + \frac{1}{2} a_2 \, t_f^2$

Solve for the two component velocities (just basic algebra) and take their Euclidean norm.

I cannot figure out why my answer is wrong. I created a right triangle with a base of $\sqrt{10^2+20^2}=10\sqrt{5}$ and a height of $10$ . The angle of inclination is given by $\theta=\tan^{-1}{\frac{1}{\sqrt{5}}}$ . The acceleration along the incline is then $g\sin{\theta}=g\sin{\tan^{-1}{\frac{1}{\sqrt{5}}}}=\frac{10}{\sqrt{6}}$ . So, the (signed) distance traveled along the incline is given by $D=vt-\frac{1}{2}g\sin{\theta}t^2=vt-\frac{5}{\sqrt{6}}t^2$ . Since the distance traveled after $10$ seconds must be the hypotenuse of the triangle, I get $10\sqrt{6}=10v-\frac{500}{\sqrt{6}}$ , which leads to $v=\sqrt{6}+\frac{50}{\sqrt{6}}$ . But this answer differs from yours.

James Wilson - 3 years, 3 months ago

Greetings. Regarding dimensionality, the problem starts out 3D, and can be reduced to 2D. It can't be reduced to 1D, because the physics take place on a planar region. Essentially, your solution assumes that the particle position can be described using a single direction vector, when it actually requires two direction vectors and two spatial coordinates (which I have called c1 and c2).

Steven Chase - 3 years, 3 months ago

Oh, I understand now. Thanks so much!

James Wilson - 3 years, 3 months ago

@James Wilson – No prob, thanks for your interest in the problem

Steven Chase - 3 years, 3 months ago

If you golf, and are hitting an uphill putt on a tilted green, you can't just hit it directly toward the hole. You have to hit it a little uphill of the hole, so it rolls back toward the hole as it moves toward it.

Great problem!

G Silb - 8 months, 2 weeks ago

Extreme Putting

The answer is 23.6758.

1 solution

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