Determine the height to which the center of the baton travels

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To determine the height the center of a baton travels when thrown straight up, the angular speed is given as 1.80 revolutions per second. The initial attempt used a kinematic equation incorrectly by assuming zero initial velocity. After recalculating with angular acceleration and the correct formulas, the height calculated was approximately 6.08 meters, which is close to the book's answer of 6.05 meters, indicating minor rounding differences. The discussion emphasizes the importance of using the correct initial conditions and equations for angular motion. Proper application of angular kinematics leads to accurate results in projectile motion scenarios.
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Homework Statement


a baton twirler throws a baton straight up in the air, it goes up and comes straight back down. it completes 4 rev. ignoring air resistance and assuming the angular speed is 1.80 rev/s , determine the height to which the center of the baton travels above the point of release.


Homework Equations



i tried to use a kinematic eq. x=1/2(vnot+v)t

3. The Attempt at a Solution [/b
i tried that equation and solved for x, i put zero for vnot and 1.8 for v. i fount t to be 2.22 seconds i got my answer to be 2m but its wrong ...thanks!
 
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1.8 is the angular velocity. Time is correct. Again initial velocity is not zero. You are using the equation in linear form so you will have use linear velocity and displacement only.
 
ok so i used the formula to find angular acceleration then used that answer and put it into this...theta=(wnot)t + 1/2 (alpha)(t^2) i got 6.08 the answer in the book is 6.05m so must just be some rounding errors. thanks!
 

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