- #1

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=.5mv

^{2}

## 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/s

^{2}

F=ma

F=(.145kg)(87.46m/s

^{2})=12.68N