# Determine vertical velocity vector on sloped surface

• hjam24
In summary, the author discusses the components of the velocity of a golf ball launched at a speed v and launch angle β on a surface with a slope of φ. The author uses the Pythagorean theorem to derive the vertical velocity component, v_v, and explains that the same logic cannot be applied to the horizontal velocity component, v_h, due to the presence of the x-component. The author then introduces a new slope, θ, across the x-axis and asks for the formula for v_v in terms of both φ and θ.
hjam24
A golf is launched at a speed v,f and launch angle, β,f. The slope of the green is equal to φ. At some point the ball is located on the rim of a hole. The side view (a) and overhead view (b) looks as in the attached image.According to the author of the [paper][2] "The Physics of Putting" the vertical velocity is given by

$$v_{f,v} = v_f \cdot \cos(\beta) \cdot \sin(\phi)$$

The horizontal velocity is equal to:

$$v_{f,h} = v_f \cdot (\cos^2(\beta_f) \cos^2(\phi) + \sin^2(\beta_f))^{1/2}$$

I understand the reasoning behind the vertical velocity component. Basically we have the hypotenuse, v,fy (v,f * cos(β,f)) and multiply this with sin(φ). Now we have the opposite. However, why is the author not following the same logic for the horizontal velocity and multiplying the hypotenuse with cos(φ)? Please show me how his formulation is derived.

My (unsuccessful) attempt:

Pythagorean theorem

$$\sin^2(\phi) + \cos^2(\phi) = \frac {v_{f,h}^2 + v_{f_v}^2} {v_f^2}$$

$$v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) = v_{f,h}^2 + v_{f_v}^2$$

$$v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_{f_v}^2$$

Substitute vertical velocity according to author

$$v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2$$

v,f is v,fy

$$v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2$$

$$v_{f,h}^2 = v_f^2\cos^2(\beta) \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_f^2\cos^2(\beta)\sin^2(\phi)$$

$$v_{f,h}^2 = v_f^2\cos^2(\beta)\sin^2(\phi) + v_f^2\cos^2(\beta)\cos^2(\phi) - v_f^2\cos^2(\beta)\sin^2(\phi)$$

$$v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi)$$

$$v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi)$$

$$v_{f,h} = v_f(\cos^2(\beta)\cos^2(\phi))^{1/2}$$

Let me know what I am doing wrong. Thanks in advance

[2]: http://www.puttingzone.com/Science/cjp-putting.pdf

#### Attachments

• seu3c.png
8.2 KB · Views: 73
Last edited:
After getting the formula of ##v_{f\ vertical}##, the author seems to have made use of Pythagoras relation
$$v_{f\ vertial}^2+v_{f\ horizontal}^2=v_f^2$$
So
$$v_{f\ horizontal}^2=v_f^2-v_{f\ vertial}^2$$

You have to consider all three dimensions. Your logic applies to the y- and z- components, but it neglects the x-component. To see this, consider the case where ##\phi=0##. Using your logic, you'd say ##v_{f,v}=0##, as expected, but ##v_{f,h}=v_f\cos\beta##. But if there's no vertical component, shouldn't ##v_{f,h}=v_f##?

A straightforward way to see what's happening is to first consider the case where ##\phi=0##. Then the components of the velocity will be
$$\mathbf v_f = -v_f \sin\beta\,\mathbf i + v_f \cos\beta\,\mathbf j + 0 \,\mathbf k.$$ Now add in the effect of the uphill slope, which mathematically is a rotation by an angle ##\phi## about the x-axis. The rotation will leave the x-component unchanged but will turn the y-component into a combination of y- and z- components.

Poster has been reminded not to generate multiple thread starts on the same question (2 threads merged)
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole. The view from above looks as follows:

Vector v can be expressed in two components, v_v and v_y:$$v_y = v \cdot \cos(\beta)$$

and

$$v_x = v \cdot \sin(\beta)$$

If we look at the surface from a different angle, we see that the surface has a slope.
The slope of the green is equal to φ (along the y-axis).

v_v is perpendicular to the y-axis. As a result:

$$v_v = v \cdot \cos(\beta) \cdot \sin(\varphi)$$

Let's assume now there is no slope across the y-axis (φ=0), but there is a slope across the x-axis, θ.

Following the same logic v_v is perpendicular to the x-axis. As a result:

$$v_v = -1 \cdot v \cdot \sin(\beta) \cdot \sin(\theta)$$

What is the formula of v_v expressed in terms of both φ and θ? So, in the case that the surface is sloped/inclined across both the x- and y-axis?

[Moderator's note: moved from a technical forum.]

Last edited by a moderator:
hjam24 said:
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole. The view from above looks as follows:
@hjam24 -- Why did you repost this question and show no work? Were the replies above to your original thread a couple weeks ago not helpful enough?

berkeman said:
@hjam24 -- Why did you repost this question and show no work? Were the replies above to your original thread a couple weeks ago not helpful enough?
Hi,

The previous question was with a singe slope across one axis, this question involves a slope across bot the x-axis and the y-axis. I believe the difference is large enough to open a new topic. I am sorry if I misjudged that

Well the wording was so similar that it was hard to tell the difference.

hjam24 said:
What is the formula of v_v expressed in terms of both φ and θ? So, in the case that the surface is sloped/inclined across both the x- and y-axis?
Please show your work on this, and you should continue to get good help. Thank you.

hjam24
hjam24 said:
why is the author not following the same logic for the horizontal velocity and multiplying the hypotenuse with cos(φ)?
In the direction of the ball, the slope is not ##\phi##.
hjam24 said:
$$v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2$$

v,f is v,fy

$$v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2$$
I don’t understand what you did there. You seem to have just multiplied a term by ##\cos^2(\beta)##.

hjam24 said:
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole.
What is your question for the second case?
The vertical velocity vector is perpendicular to both top views, as well as to the axis that is not tilted (or remains horizontal).

It would be nice if the OP actually replied to the posts so we know if he understood what was said or not.

Lnewqban

## 1. How is the vertical velocity vector on a sloped surface determined?

The vertical velocity vector on a sloped surface is determined by using the formula v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time elapsed. This formula takes into account the slope of the surface and the acceleration due to gravity.

## 2. What factors affect the determination of the vertical velocity vector on a sloped surface?

The determination of the vertical velocity vector on a sloped surface is affected by the slope of the surface, the initial velocity, the acceleration due to gravity, and the time elapsed. Other factors such as air resistance and friction may also have an impact on the vertical velocity vector.

## 3. How does the vertical velocity vector change as the slope of the surface changes?

The vertical velocity vector changes as the slope of the surface changes because the acceleration due to gravity is affected by the slope. The steeper the slope, the greater the acceleration due to gravity and therefore the greater the vertical velocity vector will be. Similarly, a shallower slope will result in a smaller vertical velocity vector.

## 4. Can the vertical velocity vector on a sloped surface be negative?

Yes, the vertical velocity vector on a sloped surface can be negative. This occurs when an object is moving downwards on a sloped surface. In this case, the initial velocity may be positive, but the acceleration due to gravity will cause the vertical velocity vector to become negative.

## 5. How can the determination of the vertical velocity vector on a sloped surface be applied in real-world situations?

The determination of the vertical velocity vector on a sloped surface can be applied in real-world situations such as calculating the speed of a skier going down a slope or the velocity of a ball rolling down a hill. It can also be used in engineering and construction to determine the forces acting on a structure on a sloped surface.

• Introductory Physics Homework Help
Replies
55
Views
3K
• Introductory Physics Homework Help
Replies
15
Views
230
• Introductory Physics Homework Help
Replies
5
Views
662
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
33
Views
4K
• Introductory Physics Homework Help
Replies
13
Views
1K