Physics HW- Falling Bodies, Part 3

[haruspex]

It's not an online test, it's just a homework set, but my physics teacher usually assumes we'll round instead of leaving answers as a square root, as far as I know.This problem could be different. He hasn't explained surds yet, or gotten into leaving answers as square roots, so that may be where some of the confusion is coming from. Would you mind clarifying what equation I need to use at this point?

CAF123's point is that if you evaluate some variable as a number, then use that number in further calculations, then any rounding error gets carried into the second calculation. Sometimes this can result in a much bigger error in the final answer than you might expect. It's almost always best to keep the algebraic form as long as you can along the path to each result. Sometimes you get some cancellation.
'surd' means any irreducible expression involving an Nth root. It comes from the Latin surdus (deaf), which is also the origin of 'absurd'. It's like calling something 'dumb'.

 

[Medgirl314]

Oh! I thought that the final answer would have to be left as a square root. Thanks for the explanation, that's neat! I expected the word was Latin, but wasn't sure for what.

 

[Medgirl314]

I'm sorry, I thought I replied to this. I have to admit, I have gotten a little lost with this problem. Would someone please help me with beginning the next step? Am I suppossed to begin the problem again with an unrounded answer? If so, what equation should I use? Should I leave my inital velocity answer the same for that part of the problem? Thanks so much! :-)

 

[Medgirl314]

Could I use vavg=distance/time, but using my unrounded answer for velocity?

 

[Medgirl314]

That seems to yield an answer of approximately .27 seconds, which seems far more plausiable than my original answer.

 

[haruspex]

That seems to yield an answer of approximately .27 seconds, which seems far more plausiable than my original answer.

No, that's much too low. Pls post your working for that.
Most of this thread has been about finding the launch speed (plus a few rabbit holes). For the time, you almost nailed it in your OP. You just forgot to take the square root.

 

[Medgirl314]

Okay, thank you! Thanks!

I used this formula: vavg=distance/time,

Plugged in my numbers:11.86591758=44/t
And got about .27.

The square root of 55 is approximately 7.42. Is that the answer?

 

[haruspex]

I used this formula: vavg=distance/time,

Plugged in my numbers:11.86591758=44/t
And got about .27.

Two problems there. The average speed is only half the launch speed, so 5.9329587897.
You seem to have divided the wrong way: you want t = 44/average speed. Correcting these will give you the same number as below.

The square root of 55 is approximately 7.42. Is that the answer?

Not quite. What time do you think you have calculated here? What time does it ask for?

 

[Medgirl314]

It asks for the total time, and that time is just how long it would take to go up. So the answer would be more like 14.84 seconds.

 

[haruspex]

It asks for the total time, and that time is just how long it would take to go up. So the answer would be more like 14.84 seconds.

Exactly.

 

[Medgirl314]

Great, thank you!