Physics help needed right now

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EditThe following data table gives the position of an object during freefall. The time interval between adjacent points is 1/60 seconds.  

Point Position (cm) 
1 94.30 
2 91.53 
3 88.48 
4 85.17 
5 81.58 
6 77.72 
7 73.58 
8 69.17 
9 64.49 
10 59.54 


What is the instantaneous velocity (in cm/s) of the object when it is at position 3?

PLEASE HELP
Sep 28th, 2015

Thank you for the opportunity to help you with your question!

instantaneous velocity (in cm/s) of the object when it is at position 3 = displacement/time=| 91.53-88.48 |/1/60 =3.05/.016667 cm/sec = 183 cm/sec.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 28th, 2015

Thanks for your help! I got that answer as well but for some reason my computer says its not right.. any other ideas?

Sep 28th, 2015

I think since it is a free fall we should use another formula, V= Vo  + -gt^2 where g=-9.8m/sec^2, wiill get back to you, in a few mins.

Sep 28th, 2015

thank you so much!

Sep 28th, 2015

Did you find another answer?

Sep 28th, 2015

V= Vo +gt

Vo= (94.30-91.53)/1/60 =2.77*60=166.2 cm/sec

g=980cm/sec^2

t=2/60 sec

V= 166.2cm/sec +  980/30 cm/sec =198.87 cm/sec

I hope this is correct this time, please let me know.

Sep 28th, 2015

it still did not work 

Sep 28th, 2015

I used t= 2/60 lets try 1/60, V=182.53

Sep 28th, 2015

that didn't work either.. i don't know why none of these are working, i only have one attempt left though.

Sep 28th, 2015

let me do some research

Sep 28th, 2015

I think I get it now, Vo  is at number 1 and is =0, At number 3, V= gt= 980cm/sec^2 * 2/60 sec=32.6667cm per 1/60 of a second . Velocity in cm per second would be 32.6667*60 = 1960cm/sec

Sep 28th, 2015

I am sure this is correct, I checked and found that acceleration is constant all through out =980cm/sec^2

Sep 28th, 2015

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Sep 28th, 2015
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Sep 28th, 2015
Oct 17th, 2017
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