Physics HW- Falling Bodies, Part 3

[Medgirl314]

Is that correct so far?

 

[haruspex]

Is that correct so far?

Yes - sorry, I didn't think you needed that verified. Numerically it is clearly correct.

 

[Medgirl314]

Okay, thanks! I wasn't sure if I plugged it in correctly, because it looks too easy to solve. I'll simplify and respond ASAP!

 

[Medgirl314]

70.8 m/s^2=70.4 m.s^2

I realize that's a bit off, but I *think* it's just because of my rounding earlier. Is the answer simply approximately 70.6 m/s? Thanks again for all your help!

 

[Medgirl314]

Sorry, I meant 70.6 s.

 

[haruspex]

70.8 m/s^2=70.4 m.s^2

I realize that's a bit off, but I *think* it's just because of my rounding earlier. Is the answer simply approximately 70.6 m/s? Thanks again for all your help!

I think we've gone a bit off track here. First, let me correct the units above, which I didn't think to do before. Both of those numbers are in units of m²/s². On one side that's from multiplying two speeds, on the other from multiplying an acceleration by a distance. This is not a route to calculating time.
This avenue seems to have started as a result of your post #31:

someone else suggested we calculate the average

and my post #32:

The reference to average velocity was distance = v_avg * time. We can rewrite [v²=v0²+2aΔx] using an average: v_avg * Δv = aΔx.

I did not mean to suggest this was a useful way to calculate the time. I was just explaining a connection between the equations.
Part b here (taken in isolation) is exactly like your jumper problem. You know the height reached and the acceleration, and you want to find the time in the air. Much the simplest way is to recognise that the time up and the time down will be the same, so we can just work out the time down and double. That let's us use the familiar equation:
s = v₀t + at²/2
where v₀ = 0 (top of jump), s = height, a = g. Solve to find t.
If you don't like that approach then you need to get an equation like the one above but using v_f instead of v₀. We know v_f = v₀ + at, so we can use that to get rid of v₀:
s = (v_f-at)t + at²/2 = v_ft - at²/2

But in this thread, part a already got you to find the initial velocity, so we can take a shortcut. You know the initial velocity and the velocity at the top (0). Because it is uniform acceleration, the average velocity is simply the average of those two. Now you can use s = v_avg*t to find the time to reach the top. Again, you need to double for the total time in the air.

I hope this undoes the confusion.

 

[Medgirl314]

Okay, thank you! So what we're saying here is that I need to use this equation:s = (vf-at)t + at2/2 = vft - at2/2 , correct?

 

[Medgirl314]

Or maybe this one, rather? s = v0t + at2/2 Only, I would have to rearrange it in order to solve for t, correct?

 

[Medgirl314]

I think I was focusing too much on the first 3/4 of your latest post. It seems that you are saying this equation would be best:s = vavg*t

My average velocity would be 5.95.

44 m=5.95 m/s^2*t

After taking the square root of both sides: 6.6 m=2.44*t

Dividing both sides by 2.44= 2.7 s

Doubling to find the total time: 5.4 s. This answer seems to make much more sense than 55 or 70-something seconds. Is that my final answer? Thanks again!

 

[CAF123]

Why did you take the square root? Also, your value of v_avg has been rounded and so you will not obtain the correct result for t. This is what caused the mismatch earlier.

However, you already calculated the time to rise before did you not? (If you understand why the time back down is the same, then to get the total time, simply multiply this by 2).

 

[Medgirl314]

I think I got this problem mixed up with another one where I needed to take the square root. Would you mind explaining when I incorrectly rounded vavg? My physics teacher says he's not too worried about how I round, because he's more worried about how I get to the answer. I understand that the time back down is the same, but since I've made this far more complicated than it needs to be, I didn't realize we already found the time.

Thank you!

 

[CAF123]

Would you mind explaining when I incorrectly rounded vavg? My physics teacher says he's not too worried about how I round.

You found the initial velocity of the rock to be 11.9m/s and so you then said that v_avg= 11.9/2 = 5.95m/s. But the value of 11.9m/s has been rounded, so to obtain the correct average velocity you should work with the exact result.

The rounding was why when we used the quadratic formula in #8 to find the time to rise, we ended up with two solutions (which in itself does not make sense), neither of which corresponded to the correct result.

In general, it is always best to not round results until the very end of a question. Work with exact results in your calculator or write out the surd e,g in the above instead of working with 11.9 work with v₀ = √2*1.6*44 = √140.8, taking the +ve sqrt since upwards was defined positive.

 

[Medgirl314]

Okay, I think I understand what you're saying. You're saying that after I calculated the inital velocity, I should have kept the entire decimal answer, and then worked through our equations without rounding, but then round at the very end, correct? So I should go back to calculating the intial velocity, record the unrounded answer, and then go back and re-do the formula in #8.

The only thing I'm not clear on is the abbrevations. What is surd and +ve? Thanks again!

 

[CAF123]

Okay, I think I understand what you're saying. You're saying that after I calculated the inital velocity, I should have kept the entire decimal answer, and then worked through our equations without rounding, but then round at the very end, correct?

Yes, and the 'entire decimal answer' is exactly √140.8.

So I should go back to calculating the intial velocity, record the unrounded answer, and then go back and re-do the formula in #8.

Is this for a online test? It might say 'Find the initial velocity to X amount of significant figures', in which case you should do so. But for subsequent parts of the same question, always use the exact value.

Convince yourself that using 11.9 in place of √140.8 you end up with inconsistent results.

The only thing I'm not clear on is the abbrevations. What is surd and +ve? Thanks again!

Sorry, √140.8 is a surd since you cannot simplify it further. It is not an abbreviation, just a common term. +ve means positive.

 

[Medgirl314]

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?

 

[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!