Kinematics of rolling ball question

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SUMMARY

The kinematics problem involves a ball rolling off a table 1.225 m high with an initial horizontal velocity of 6 m/s. To determine the time it takes for the ball to hit the ground, the vertical motion should be analyzed independently of the horizontal motion. The correct approach is to use the equation for vertical displacement, where the initial vertical velocity (Vo) is 0 m/s, leading to the formula Distance = Vo * t + 1/2 * a * t^2. This results in a time of approximately 0.5 seconds for the ball to reach the ground.

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  • Familiarity with gravitational acceleration (9.8 m/s²)
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  • Learn to apply kinematic equations for vertical and horizontal components
  • Explore the effects of initial velocity on projectile trajectories
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Homework Statement



Hello, new here. I'm doing homework for my Honors Physics class, and I have a kinematics problem that seems simple enough, but I just can't get it. Help, please?

A ball is rolling on a table 1.225 m high at 6 m/s. After it rolls off, how long does it take to hit the ground?


Homework Equations



I used V^2= (Vo)^2 + 2a(X-Xo) to find final velocity, than used V= Vo+at to find time.


The Attempt at a Solution



V^2= (6 m/s)^2 + 2(9.8 m/s^2)(1.225 m) V= 7.7 m/s

t= (V-Vo)/a = (7.7 m/s - 6 m/s)/ 9.8 m/s^2 t= .17 s

I know this is wrong. My physics teacher told me to use Vx, not Vy, for Vo. I'm not quite sure how to go about this. Do I need to find the angle (which I'm not sure how to do in this problem) and use Vox= Vocos(theta)?

Any help would be appreciated. :redface:
 
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On the table the ball only had a horizontal velocity... that only affects horizontal displacement... it doesn't affect vertical displacement.

What is the initial vertical velocity? You have a vertical displacement of -1.225m.
 
It doesn't matter what the horizontal velocity is, the 6m/s is there to confuse you. Only consider the vertical motion, Use Distance= Vo + 1/2at^2 where Vo (the initial vertical velocity) is 0. Oh, and only consider the distance traveled to be negative if you consider the acceleration to be negative.
 

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