1. The problem statement, all variables and given/known data A baseball is thrown at 85 mph and is thrown with a spin rate of 125 rpm. The distance between the pitchers point of release and the catcher’s glove is 60.5 feet. How many full turns does the ball make between release and catch? 2. Relevant equations Vf^2=V0^2+2a(Δ x) X= V0t+1/2at^2 ω=ω0+αt ΔΘ=ω0t+1/2αt^2 3. The attempt at a solution I converted.. 60.5 ft (1m/ 3.280 ft)= 18.45 meters 85 mph (1609.34 meters/1 mile)(1 hour/3600 s= 38 m/s 125 rev/min(2pi/1 rev)(1 min/ 60 s) =13.08 rad/s I think we would have to use the velocity and distance to find the acceleration and then the time... So, I used.. Vf^2=V0^2+2a(Δ x), I assumed I had to set the initial velocity to 0 because the ball would be at rest prior to being thrown..and so the initial angular velocity to 0 also.. 38 m/s= 0 + 2a(18.45) A= 1.03 m/s^2 Then.. X= V0t+1/2at^2 18.5 m= 0+1/2(1.03)(t^2) T=5.99 s But, I think something is most definitely wrong within that. Because then when I plug the time into.. ω=ω0+αt 13.09rad/s=α(5.99s) α=2.19rad/s^s ΔΘ=ω0t+1/2αt^2 =1/2(2.19rad/s^2)(5.99s)^2 ΔΘ=39.29 rads=6.25 revolutions, which is not the answer. I'm not sure if I did the entire thing wrong or where it went south.. or if i'm complicating the problem too much. But, I don't know how else to solve for this without finding the linear acceleration and then for time, then using it for rotational motion..