EditThe following data table gives the position of an object during freefall. The time interval between adjacent points is 1/60 seconds.
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.
Thanks for your help! I got that answer as well but for some reason my computer says its not right.. any other ideas?
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.
thank you so much!
Did you find another answer?
V= Vo +gt
Vo= (94.30-91.53)/1/60 =2.77*60=166.2 cm/sec
V= 166.2cm/sec + 980/30 cm/sec =198.87 cm/sec
I hope this is correct this time, please let me know.
it still did not work
I used t= 2/60 lets try 1/60, V=182.53
that didn't work either.. i don't know why none of these are working, i only have one attempt left though.
let me do some research
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
I am sure this is correct, I checked and found that acceleration is constant all through out =980cm/sec^2
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