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Phillipv2004
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Homework Statement
The average force needed to throw a baseball 90mph. I tried to find the force at 90mph on release and 90mph at the plate and got two wildly different answers and it seems off that they vary soo much.
Homework Equations
· Velocity- 90 mph or 40.23 m/s measured at release point.
· Mass- Objectives of the Game. Rules of Baseball. Major League Baseball Enterprises, 1998."It shall weigh not less than five nor more than 5 ¼ ounces avoirdupois" – 142g-149g so 145g=.145kg
· Distance- Ball accelerated over a distance of 1.75m arm length
Work= Force x Distance
Force= Work/Distance
KE=.5mv2
The Attempt at a Solution
KE=1/2(.145kg)(40.23m/s)2
KE=117.34J
F=117.34J/1.75m=67.05N
I then attempted to determine a 90mph baseball at the plate.
a=ΔV/Δt
Major League Baseball the mound and plate are 18.44m apart
t=18.44m/40.23m/s=.46s
a=(40.23m/s-0)/.46s
a=87.46m/s2
F=ma
F=(.145kg)(87.46m/s2)=12.68N