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course phy 121

Report the resulting speeds in the box below, three speeds for each setup and ramp orientation, in the same order and the same format used in the preceding:For the setup you skipped, just enter 'skipped'.

---------->>>>> ball speeds based on mean distances

Your answer (start in the next line):

14.52cm/0.05 s= 290 cm/s

18.9cm/0.05s= 378 cm/s

23.98cm/0.05 s= 479.6 cm/s

14.46 cm/0.05 s=289.2cm/s

21.08cm/0.05 s=421.6 cm/s

24.98 cm/ 0.05 s= 499.6 cm/s

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For 0.05 sec I used vf^2=vi+2a*'ds

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You still haven't said what you used for a , vi and `ds. So I can't tell what you've done wrong.

.05 sec, however, would correspond to a drop of less than 2 cm. That is something you would be able to check, to verify your results for `ds.

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vf=44.27 cm/s

Then I took (vf-v0)/a=‘dt

(44.27 cm/s-0)/980 cm/s^2= 0.05 sec

????????????????????? I am still struggling??????????????????"

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Your .05 sec result is not documented, in the sense that in your solution for `dt you haven't connected it to your data.

However .05 sec is clearly incorrect.

If the .05 s was correct, your calculations of vAve for the horizontal motion would be correct, and you would have the final velocities on the ramp.

So if you can correct the .05 s, you should be able to complete the analysis.
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@& You still haven't said what you used for a , vi and `ds. So I can't tell what you've done wrong. .05 sec, however, would correspond to a drop of less than 2 cm. That is something you would be able to check, to verify your results for `ds. *@

@&
You still haven't said what you used for a , vi and `ds. So I can't tell what you've done wrong.

.05 sec, however, would correspond to a drop of less than 2 cm. That is something you would be able to check, to verify your results for `ds.

*@