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Ball Rolling Off of Table 
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#1
Oct2509, 09:21 PM

P: 11

1. The Lab
The ball reaches the end of the table in .178584 s, a distance of 14.016 cm. The table is 86.49 cm tall. What is the ball's velocity and far will it roll? 2. Relevant equations y=v_{i}t+.5(g)(t)^{2} x=vt 3. The attempt at a solution I assumed v_{i} was 0 and got that time=4.201, but that didn't seem to make sense. Should I convert the 86.49 into .8649 m? Because then I get that time=.4 ish and that can't be right.... Edit: Do I multiply the time (4.20) by vhorizontal? 4.20*(14.016/.1785)=32.98428? 


#2
Oct2609, 02:12 AM

P: 1,395

The question is rather unclear. Is the velocity when the ball reaches the ground meant, or when it leaves the table?. Does "and far will it roll?" mean how far from the table will it land?
You certainly need to convert the distance to meters to get the falling time. Why is .4ish not right for the time? You do need to multiply the falling time with the horzontal speed to get the distance from the table. Watch the units. To answer the velocity question you need to include both horizontal and vertical speed. 


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