What is the average drag force on a softball pitched at 99.6 km/hr?

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Homework Help Overview

The problem involves calculating the average drag force on a softball pitched at a speed of 99.6 km/hr, considering a mass of 0.250 kg and a 10.8 percent reduction in speed by the time it reaches the plate, which is 15.2 m away. The discussion centers around the application of physics principles related to forces and motion, specifically air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting velocity from km/hr to m/s and calculating the final velocity after accounting for the speed reduction. They explore using kinematic equations to find acceleration and apply Newton's second law to determine the drag force. Questions arise regarding the interpretation of "average force" and whether it should be calculated over distance or time.

Discussion Status

Some participants have provided calculations and feedback on each other's work, with one confirming the correctness of another's approach. There is an ongoing exploration of the implications of averaging methods for force calculations, and one participant has noted a discrepancy in the sign of the final answer, indicating a productive exchange of ideas.

Contextual Notes

Participants are operating under the assumption that gravity can be neglected, and there is a focus on the definitions and implications of average force in the context of the problem. The discussion reflects a mix of interpretations regarding the calculation methods available based on the information provided.

kopinator
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A softball having a mass of 0.250 kg is pitched at 99.6 km/hr. By the time it reaches the plate, it may have slowed by 10.8 percent. Neglecting gravity, estimate the average force of air resistance during a pitch, if the distance between the plate and the pitcher is 15.2 m.


F=ma
Fdrag=(1/2)pACv^2 p=density, A=cross sectional area, C=drag coefficient
Vf^2=Vi^2 + 2a(X-xi)


I converted the velocity into m/s and and then found the final velocity as it reached 15.2 m and plugged them into the 3rd equation to find the acceleration. After that I used F=ma knowing the acceleration and mass. This still wasnt right.
 
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kopinator said:
I converted the velocity into m/s and and then found the final velocity as it reached 15.2 m and plugged them into the 3rd equation to find the acceleration. After that I used F=ma knowing the acceleration and mass. This still wasnt right.
Sounds right to me. Show the details of your calculations.
 
99.6 km/h=27.67 m/s
27.67 x 10.8%=2.98836
27.67(Vi) - 2.98836=24.68 m/s approx.(Vf)
(24.68^2)=(27.67^2) + 2a(0-15.2) solve for a
a=5.148898026 or 5.15 m/s^2
F=ma
F=(.250kg)(5.15 m/s^2)=1.29 N
 
I'd say that your work and your answer look correct.
 
What does it mean to say "average force"? It could be an average over distance, ∫F.ds/Δs, or an average over time, ∫F.dt/Δt. If the force is not constant, these will generally produce different answers.
To me, average over time is the more natural interpretation. It certainly fits better with the usual meanings of average velocity and average acceleration. A well written question tells you which is meant; some questions only give you enough information to calculate one; here (using the drag formula quoted in the OP) you have enough for both.
It only makes a tiny difference here. I get 1.2835N, compared with 1.2862 for average over distance. But rounding to 3 sig figs that's 1.28 instead of 1.29, so maybe that's enough.
 
I figured it out. I used the longer way to find the answer using W=deltaK. The answer is -1.29 N. I had the right number but not the right sign. Thanks for your help!
 

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