A baseball approaches home plate at a speed of 42.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 51.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.70 ms. What is the average vector force the ball exerts on the bat during their interaction

F= 3582j + 4350i

Explanation:

Given data

Let the average vector force be F

the initial and final momentum be Pi and Pf respectively

Velocity v1=42m/s

and v2=51m/s

Mass m=145g=0.145kg

Time t=1.7ms=0.0017s

We know that

F= pf - pi

The baseball initially moves in the –x direction

F= mv1 – (-mv2)

F = [mv1– (-mv2)] / t

F= [(0.145)(42.0)j + (0.145)(51.0i)] / 0.0017

F= 3582j + 4350i

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.