A baseball pitcher pitches a fastball with a horizontal velocity of 40 m/s. The

Question

A baseball pitcher pitches a fastball with a horizontal velocity of 40 m/s. The horizontal distance from the point of release to home plate is 17.50 m. The batter decides to seing the bat 0.30 s after the ball was released by the pitcher. The average angular velocity of the bat is 12 rad/s. The angular displacement of the bat from the batter's shoulder to hitting positions above the plate is between 1.5 and 1.8 rad.

a. Will the bat be in a hitting position above the plate when the ball is above the plate? Assumme the pitch is in the strike zone.

Explanation / Answer

First we apply d = vt to find out how fast the ball is over the plate

17.50 = 40(t)

t = .4375 sec

Now we can find out how long it will take the bat to travel to the 1.5 to 1.8 rad distance

1.5 = (12)t

1.8 = (12)t

t = .125 sec to get to the 1.5 rad location

t = .15 sec to get to the 1.8 rad location

Add the additional time of the 0.3 sec the batter waited to swing and we see that the bat will be in the hitting zone between

.425 and .450 sec

Since the ball will be in the hitting zone at .4375 sec, that is in between the time required, so yes, the bat is in a hitting position when the ball is above the plate.

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