A baseball player extends his arm straight up to catch a 145-g baseball moving h

Question

A baseball player extends his arm straight up to catch a 145-g baseball moving horizontally at 42 m/s. It's 63 cm from the player's shoulder joint to the point where the ball strikes his hand, and his arm remains stiff while it rotates about the shoulder during the catch. The player's hand recoils nearly horizontally while he stops the ball in 2.5 ms (1 ms = 1/1000 s). What average torque does the player's arm exert on the ball? Check the magnitude of your answer to 2 s.f. The ball had lots of kinetic energy before it was caught. In your solution calculate how much kinetic energy the ball had and explain where all that kinetic energy went. Define your axes and use vectors

Explanation / Answer

mass of the base ball . m=145 g

speed, v=42 m/sec

arm length, l=63 cm

time, t=2.5*10^-3 sec

a)

torque=F*l

=m*a*l

=m*(v/t)*l

=145*10^-3*(42/(2.5*10^-3))*(0.63)

=1.5*10^3 N.m

b)

K.E=1/2*m*v^2

=1/2*145*10^-3*(42^2)

=127.89 J

the amont of kinetic enegry convert into rotational energy of the players hand

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