a basball pitcher pitches with a horizontal velocity of 80mph(36m/s) the horizon

Question

a basball pitcher pitches with a horizontal velocity of 80mph(36m/s) the horizonatal distance from the point of release to home plate is 59ft. the batter decides to swing thebat 0.30seconds after the ball was released by the pitcher. the batter swings the bat with an average angular velocity of 675 degree/second. the angular displacement needed to get the bat from behind the batter's shoulder to the hitting position above home plate is 155degree

now assuming that the batter does hit the ball and makes contact at the sweet pot. the bat has an intantanous angular velocity of 1375.2 degree/second at the instant of contact. the distance from the axis of rotation for the swing to the sweet spot of the bat is 1.25meters.

in m/s what is the intantous linear velocity of the sweet spot at the instant of ball contact?

Explanation / Answer

Two steps:

1) The horizontal velocity is constant. The first question you have to ask is, "how long did it take to get to home plate?"

Formula:
distance = velocity x time.
If velocity is 36 m/s, and distance is 59? (that seems wrong). Either way, solving for t gives you:

t= distance/velocity. Solve for t.

2) How far did the ball travel in the vertical direction during that time period?

formula:

Distance = (initial vertical velocity) x t + (1/2)(acceleration)(t^2).

In this case, the initial vertical velocity is 0. the value of 't' is the same value you solved for above. The acceleration is due to gravity which is -9.8m/s^2. Plug those values in and you have your answer.

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