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

Question

A baseball pitcher pitches a fastball with a horizontal velocity of 40m/s. The horizontal distance from the point of release to home plate is 17.5m. Jacque decides to swing the bat 0.3s after the pitcher released the ball. The average angular velocity of the bat is 12 rad/s. the angular displacement of the bat during the swing is between 1.5 rads and 1.8 rads. (use 1.65 rads).

A. Will jacquis bat be in the hitting position above the plate when the ball crasses the plate?(assume the pitch is in the strike zone)

B. Assume Jacqui does hit the ball. if her bats instantaneous angular velocity is 30 rad/s at the instant of contact , and the distance from the axis of rotation to the sweet spot of the bat is 1.25m, what s the instantaneous linear velocity of the sweet spot at the instant of ball contact?

Explanation / Answer

Distance d = Vt

time t= d/V = 17.5/40

t = 0.437 secs

time available t = 0.437 - 0.3 = 0.137 secs

angle mioved theta = 0.137 *12 = 1.65 rad

so bat will be in position

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B) linear velocity is V=r*W

V = 1.25 * 30

V = 37.5 m/s

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