Help Solve Golf Launch Speed Problem w/o Initial Velocity or Time

  • Thread starter ConfusedStudent
  • Start date
In summary, the problem discussed was related to finding the launch speed of a golf club thrown at an angle of 30 degrees above the horizontal, given that it lands 50 m away. Various equations were suggested, including the range formula and kinematic equations, and it was determined that the mass of the club is not relevant. Ultimately, the correct solution was found to be 23.8 m/s using the equation V0 = square root of -xay/(2cosθsinθ).
  • #1
ConfusedStudent
11
0
Help with a problem!

This problem seems pretty easy but for the life of me I cannot get it, I think it's confusing me because I don't have a intial velocity or a time...help!

A golfer hurls his golf club across the green. Launched at a angle 30 degrees above the horizontal, the club lands 50 m away, what was the launch speed of the 4 kg club?

I've tried this problem over and over and I can't get it, as far as I can tell the mass is useless right?

The way I tried it before was by using one of the kinematic equations, I now know that this problem has to be split into it's x and y components, from there...I need help.
 
Physics news on Phys.org
  • #2
Yes, mass is unimportant.

Make an expression for x velocity relating time spent in flight and how far the club flew.

Make an expression for initial y velocity using time of flight and gravitaional acceleration.

Relate the expressions for x velocity and y velocity, eliminating time, using the launch angle.

Njorl
 
  • #3
Perhaps this will help, using a range formula.

We will assume no air resistance and be using g as the accleration due to gravity and sin Θ to represent the heigth of the clubs in the air (2Θ since the clubs must go up and down) we achieve the following variant of a kinematic equation:

x = Vinitial squared * sin(2Θ)/g. Solving for Vinitial = square root of gx/sin(2Θ) or 31.3 m/s

The final answer is the actual initial velocity not just the x-component.

NOTE: The range formula only works if the initial take off point and the landing point are of the same height. As the initial height of the man is unknown, you will have to assume the clubs land at the same heighth as the man (on a hill perhaps) or the man is laying on the ground when they are thrown.
 
Last edited:
  • #4
An object moving with a constant horizontal speed vx has distance function x= x0+ vx t.

An object moving with initial vertical speed vy and constant vertical acceleration -g has distance function y= y0+ vy t- (1/2)g t2.

In this problem you can take x0 and y0 to be 0. The club hits the ground when y= 0 (again) so that gives you t. Put that into
x= x0 + vx t= 50 to find the initial speed.
 
  • #5
Game Guru , I am correcting a test I have already had so I have it to study, and your answer is what I put on the test and got a 0/15...so it's not that.

This is what I'm thinking,
I use the Range formula:
range=initial velocity squared divided by g, and that multiplied by sin 2 theta,

or 50=(Vo^2/9.8) *sin 2(30 degrees)which gives me 23.8 m/s...Am I close? Anybody?

I still can't figure out how to do this when all the equations leave me which both time and initial velocity missing.
 
  • #6
My apologies ConfusedStudent. Although the range formula is correct, my initial calculation may have been done in error. When double-checking today, I do get 23.8 m/s as V0.

Here is another way of solving the problem.

The horizontal displacement of the clubs is given by x=V0*t+at2/2 with a = 0: x=V0 * t = (V0 cosθ)t. The vertical component of the club's velocity at any time t is given by Vy = Vinitialy + ayt. At the instant that the clubs land, vy=-v0y. Therefore, the vertical component of the club's velocity is

-v0y=v0y + ayt

If you solve for t (up being positive):

t = -2voy/ay = -2v0sinθ/ay

Plug this expression for t into the equation for x gives you

x=(v0cosθ)t = v0cosθ(-2v0sinθ/ay)

or in more simpler terms:

x=-2v02sinθcosθ/ay

Solving for V0 = Square root of -xay/(2cosθsinθ) = 23.8 m/s.

Hope that makes it clearer and sorry for my late night miscalculation.
 
Last edited:

1. What is the issue with the current method of measuring golf launch speed?

The current method of measuring golf launch speed relies on initial velocity and time, which can be affected by external factors such as wind or human error. This can lead to inaccurate measurements and inconsistencies in data.

2. Can't the issue be solved by using more accurate equipment?

While using more accurate equipment can help improve the accuracy of measurements, it does not address the root problem of relying on initial velocity and time. Additionally, upgrading equipment can be costly and may not be feasible for all golfers.

3. Is there a scientific solution to this problem?

Yes, there are scientific methods that can be used to calculate launch speed without relying on initial velocity and time. These methods involve using principles of physics and motion to determine the launch speed of the golf ball.

4. How does the scientific solution work?

The scientific solution involves using equations such as the law of conservation of momentum and the equations of motion to calculate the launch speed of the golf ball. This method takes into account factors such as the mass of the ball, the angle of the club face, and the spin of the ball.

5. Will the scientific solution be adopted by all golfers and professionals?

It may take some time for the scientific solution to be adopted by all golfers and professionals, as it may require a shift in mindset and a learning curve. However, as more research and data are gathered to support its accuracy, it has the potential to become the standard method for measuring golf launch speed.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
16
Views
974
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
17
Views
365
  • Introductory Physics Homework Help
Replies
2
Views
818
  • Classical Physics
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
535
Back
Top