A baseball player hits an inside fastball at an angle of 25

A baseball player hits an inside fastball at an angle of 25 degrees with the horizontal and the baseball travels 450 feet on the horizontal field. What is the average impulsive force the bat imparts to the baseball if the bat maintains contact for 0.4 milliseconds? The baseball weighs 5.25 oz.

Solution

distance d = 450 feet = 137.16 m

mass m = 5.25 oz = 0.148 kg

time t = 0.4 millisecond

angle 25 degree

square root = sqrt

horizontal impulsive force vh

vertical impulsive force vv

average impulive force= sqrt(vh^2=vv^2)

initial velocity u = 0

final velocity v =2d/t-u

2*137.16/0.4

6.85 m/s

force on horizontal direction = mv-mu

=0.14*6.85cos25-0

0.869m/s

force on vertical direction

= mv-mu

=0.14*6.85sin25-0

=0.405m/s

average impulsive force

Ft=sqrt(vh^2+vv^2)

sqrt(0.869^2+0.405^2)

=0.958Ns

average impulsive force= 0.958 Ns