Calculating Force Applied to Golf Ball by Club with Limited Data

• Nevolute
In summary, the golf club forces the ball to travel a distance of 250m and to attain a speed of 61.74 m/s.
Nevolute
Homework Statement
You hit a golf ball (mass 50g) squarely with the club face from ground level on a flat golf course. The ball leaves the club at an angle of 20 degrees above horizontal and travels 250m before hitting the ground. The force vs time graph below shows the force exerted on the ball by the club. Find the maximum value of the force applied to the ball by the club. Ignore air resistance.

The attempt at a solution
I feel like we are missing one variable. So we know the distance the ball travels but no initial speed. I looked at kinematics equations but I can't solve because we are not given any speed or time (aside from the time of the impulse). I can't use energy equations because we don't know h for Potential Energy = mgh. Can't use Kinetic Energy =1/2mv^2 because we don't know v. Can't use any of the kinematics equations because we are missing either time or v initial or delta y.

Basically, the graph gives us the time it takes for the change in p (momentum represented as mass and velocity) to occur. The initial momentum of the golf ball is 0 and the final momentum can be known only with velocity, which is a consideration of distance and time.

We know how far the ball travels but not the time it spends in the air nor the height it attains (can't use trig with the angle since the motion is projectile like and thus parabolic). If we knew how hard the ball was struck we could calculate a velocity or if we knew either the time or height attained by the ball, we could calculate a speed to find the momentum/force imparted to the ball be the club. I don't think this problem is solvable with the current data.

Anyone have any insight into this?

Hello, and welcome to PF!

A couple of hints:

You'll need to know how the impulse of a force is related to the graph of the force vs time.

You have enough information about the projectile motion of the ball to determine the speed that the ball leaves the club. Remember to separate the horizontal and vertical components of the motion. For the horizontal motion, use the given horizontal range. For the vertical motion, use the fact that the ball starts at ground level and returns to ground level.

Hello, and thank you for the warm welcome!

Couldn't I simply use the Range formula to determine the velocity?

R = (v2)(sin(2θ))/gravity, where R = 250m, solving for v.

v = 61.74 m/s

Net impulse = change in momentum.
FΔt = mv-mvο, where Δt = 4ms = 4 x 10-3 s

The impulse is also the area under the force vs. time graph (A = bh/2)

So then if I equate them, mv = (FMAXΔt)/2, and solve for FMAX?

mv = ((FMAXΔt)/2)

FMAX = (2*(0.05 kg)(61.74 m/s))/4 x 10-3 s

FMAX = 1543.5N

Edit: Math.

Last edited:
Nevolute said:
Hello, and thank you for the warm welcome!

Couldn't I simply use the Range formula to determine the velocity?
Yes, of course.

In your original post, you left out the part of the template where you are asked to list relevant equations. So, I didn't know whether or not you are familiar with the range formula. In many classes, students are expected to work out projectile motion problems using only the basic equations for constant acceleration.

R = (v2)(sin(2θ))/gravity, where R = 250m, solving for v.

v = 55.8 m/s

I get a somewhat larger value for v, but your setup looks correct.

Net impulse = change in momentum.
FΔt = mv-mvο, where Δt = 4ms = 4 x 10-3 s
OK. Here, F on the left side of the equation would be the average force during the impulse.

The impulse is also the area under the force vs. time graph (A = bh/2)]
Yes!

So then if I equate them, mv = (FMAXΔt)/2, and solve for FMAX?

mv = (FMAXΔt)2

FMAX = (2*(0.05kg)(55.8m/s))/4 x 10-3 s

FMAX = 1395N

OK. Since I got a different value for v, I get a different value for FMAX. But your method looks correct!

I just realized I was calculating with R = 200m instead of 250m!

v = 61.74 m/s when I use 250m!

So in the end, FMAX = 1,543.5N

Thank you very much!

Good Work!

Nevolute

1. How does the club's swing speed affect the distance the golf ball travels?

The speed of the club's swing has a direct impact on the distance the golf ball travels. The faster the club is swung, the more force it will impart on the ball, resulting in a longer distance traveled. However, it is important to note that factors such as the angle of the club face and the type of club used also play a role in the distance achieved.

2. What is the optimal angle for hitting a golf ball with a club?

The optimal angle for hitting a golf ball with a club is between 10-15 degrees. This angle allows for enough loft to lift the ball off the ground, while also providing enough force to send the ball flying through the air. However, the angle may vary depending on the type of shot being played and the golfer's individual swing style.

3. How does the type of golf ball affect the outcome of a shot?

The type of golf ball used can have a significant impact on the outcome of a shot. Different types of balls have varying levels of spin, compression, and dimple patterns, all of which can affect the trajectory and distance of the ball. It is important for a golfer to choose a ball that best suits their swing and playing style to achieve optimal results.

4. Why do golf balls have dimples?

Golf balls have dimples because they help reduce drag and increase lift, allowing the ball to travel farther. The dimples create a turbulent boundary layer around the ball, which helps it to stay in the air longer and travel through the air more efficiently. Without dimples, golf balls would not be able to travel as far or as accurately.

5. How does the material of a golf club affect its performance?

The material of a golf club can greatly impact its performance. Different materials, such as steel, titanium, and graphite, have varying levels of flex, weight, and durability, which can affect the speed, accuracy, and distance of a shot. It is important for a golfer to choose a club with the right material to suit their swing and playing style for optimal performance.

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