Calculating Force Applied to Golf Ball by Club with Limited Data

[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?

 

[TSny]

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.

 

[Nevolute]

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-mv_ο, 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 = (F_MAXΔt)/2, and solve for F_MAX?

mv = ((F_MAXΔt)/2)

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

F_MAX = 1543.5N

Edit: Math.

 

[TSny]

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-mv_ο, 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 = (F_MAXΔt)/2, and solve for F_MAX?

mv = (F_MAXΔt)2

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

F_MAX = 1395N

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

 

[insightful]

OP, check your math!

 

[Nevolute]

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, F_MAX = 1,543.5N

Thank you very much!

 

[TSny]

Good Work!