# Average drag force

1. Feb 17, 2013

### kopinator

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.

2. Feb 17, 2013

### Staff: Mentor

Sounds right to me. Show the details of your calculations.

3. Feb 17, 2013

### kopinator

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

4. Feb 17, 2013

### Staff: Mentor

5. Feb 17, 2013

### haruspex

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.

6. Feb 17, 2013

### kopinator

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!