Projectile Motion Golf Question

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The discussion revolves around calculating the average acceleration of a golf ball that struck a spectator's head during a tournament. The ball's initial and final velocities, along with the angles of impact and rebound, are provided. Participants emphasize the need to analyze the problem in two dimensions, suggesting the use of a diagram to clarify the angles involved. While one contributor notes that the vertical acceleration due to gravity is -9.81 m/s², others highlight the importance of considering the acceleration during the brief contact with the head. The conversation concludes with the understanding that the ball experiences changes in speed both perpendicular and parallel to the surface of the head.
michaeljf
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Homework Statement


Not too long ago, an interesting thing happened at the Canadian Open Golf Tournament. Tiger Woods bounced his tee shot off of a young spectator’s head in the last round of the tournament. The ball made contact for 0.08 seconds, and struck the boy’s head at a velocity of 32 m/s, 25degree from a tangent on the surface of his cranium. It bounced off at an angle of 22degree, with a speed of 25 m/s. What was the average acceleration of the ball as it amazingly made its way back onto the fairway?

I have no idea on how to start this.


Homework Equations





The Attempt at a Solution

 
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What's the difference between the final velocity of the ball and its initial velocity? Remember that acceleration is just change in velocity divided by change in time.
 
Welcome to PF, Michael.
Not a particularly clear question! I would be tempted to answer that the acceleration is -9.81 m/s² as it "makes its way back onto the fairway" but I suppose the question really means the acceleration while in contact with the boy's head. There are 2D directions involved so you need a diagram sorting out those angles and you need to calculate the acceleration in each of the two dimensions and then get a combined value. Just use a = Δv/Δt.
 
Delphi51 said:
Welcome to PF, Michael.
Not a particularly clear question! I would be tempted to answer that the acceleration is -9.81 m/s² as it "makes its way back onto the fairway" but I suppose the question really means the acceleration while in contact with the boy's head. There are 2D directions involved so you need a diagram sorting out those angles and you need to calculate the acceleration in each of the two dimensions and then get a combined value. Just use a = Δv/Δt.

I thought in projectile motion, the horizontal velocity does not change, thus a = 0 and its 9.81 for the vertical component?
 
That's true for the free flight (neglecting air resistance).
But my theory is that you are asked for the acceleration when the head is exerting a force on the ball. I haven't drawn the diagram, but since the angle with the head changes, I suspect the ball changes speed in both the direction perpendicular to the head and the direction parallel to it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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