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

AI Thread Summary
The discussion focuses on calculating the average drag force on a softball pitched at 99.6 km/hr, which slows down by 10.8% before reaching the plate. Participants convert the initial velocity to meters per second and apply kinematic equations to find acceleration and subsequently the drag force. The calculated average drag force is approximately 1.29 N, with slight variations depending on whether average force is interpreted over distance or time. The importance of clarity in the question regarding the definition of "average force" is emphasized, as it can lead to different results. Ultimately, the correct drag force is confirmed to be -1.29 N, indicating the direction of the force opposing the motion.
kopinator
Messages
41
Reaction score
1
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.
 
Physics news on Phys.org
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!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top