Question

# A baseball approaches home plate at a speed of 47.0 m/s, moving horizontally just before being...

A baseball approaches home plate at a speed of 47.0 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 52.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.80 ms. What is the average vector force the ball exerts on the bat during their interaction?

Let the initial direction of motion of the ball as positive x axis and vertical up direction as the positive y axis, we can write the

Initial velocity of the ball, vi = (47) m/s

Final velocity of the velocity, vf = (52) m/s

So, change in velocity of the ball = vf - vi

= (52 - 47) m/s

Thus change in momentum = m(vf - vi)

= 0.145(52 - 47)

= (7.54 - 6.815) kg.m/s

From impulse - momentum theorem, we have,

Impulse = change in momentum

=> Fav*t = m(vf - vi)

=> Fav = m(vf - vi)/t

= [(7.54 - 6.815)/1.8*10^-3] N

= (418.89 - 378.62) N

So, average vector force = (418.89 - 378.62) N

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