Physics HW- Falling Bodies, Part 3

[Medgirl314]

Homework Statement

The moon is smaller than the Earth and has a weaker gravity. On the moon's surface, the acceleration due to gravity is 1.6 m/s^2. You are on the moon and you throw a moon rock straight up. It reaches a height of 44 m.

a) Find the moon rock's initial velocity.

b)Find the total time of the moon rock's flight.

Homework Equations

v0^2=v^2-2αΔy
t^2 =2y/a

The Attempt at a Solution

a) v0^2 =0-2(1.6 m/s^2)(44 m)
√v0^2=√140.8

v0=11.9 m/s



b)
t^2 =2y/a
t^2= 88 m/1.6 m/s^2

Taking the square roots of both sides:
t=55 s.

Are these answers correct? This is a newer concept to me, so input would be greatly appreciated!

 

[CAF123]

t^2 =2y/a
t^2= 88 m/1.6 m/s^2

Taking the square roots of both sides:
t=55 s.
Are these answers correct? This is a newer concept to me, so input would be greatly appreciated!

You have neglected the fact that the rock had an initial velocity. Also, you have missed a negative in your calculation of v_o. As you wrote it, you would have the sqrt of a negative.

 

[Medgirl314]

Okay, thank you! Was first answer was correct, besides the missed negative? I thought that the two negatives canceled out, but I probably did that step wrong.

Did I use the wrong formula? I can't seem to find a better one on my equation sheet. Do you have a suggestion?

Thank you!

 

[CAF123]

Okay, thank you! Was first answer was correct, besides the missed negative? I thought that the two negatives canceled out, but I probably did that step wrong.

The important thing to note is that you are dealing with vector quantities. The kinematic equation you used was $v^2 = v_o^2 + 2as$. $v = 0$ at the apex of the throw. Set up a coordinate system with which to choose the positive direction for each vector quantity. Taking upwards as positive, you see that a = -g since gravity acts downwards. v_o and s are positive since the ball is thrown upwards. Hence you get $-v_o^2 = -2gs$ and now you see the negatives cancel. If you define positive as downwards, this is also fine.

Did I use the wrong formula? I can't seem to find a better one on my equation sheet. Do you have a suggestion?Thank you!

Not necessarily the wrong kinematic equation, but you are assuming the initial velocity is zero.

 

[Medgirl314]

I understand what you're saying, besides the part about the negative. Are you agreeing that the negatives cancel out, but I stated something wrong in my formula?

Should it look something like this?

v0^2=v^2-2aΔy
v0^2=0-2(-1.6 m/s^2) (44 m)
-v0^2= -√140.8
=11.9 m/s

So would it be the same answer, but the negatives cancel differently earlier?


Using the formula t^2=2y/a , I can't seem to figure out where the inital velocity comes into play, although it has to come in somewhere!

Thanks again!

 

[CAF123]

Should it look something like this?

v0^2=v^2-2aΔy
v0^2=0-2(-1.6 m/s^2) (44 m)
-v0^2= -140.8
=11.9 m/s

So would it be the same answer, but the negatives cancel differently earlier?

v_o²=v²-2aΔy is the same as v²=v_o²+2aΔy, just rearranged. What you have done is completely correct, but I don't see why you took the negative onto the LHS. The only reason I had a negative on the RHS in my previous post is because I started with v^2 = v_o^2 + 2as and rearranged for v_o^2.

Using the formula t^2=2y/a , I can't seem to figure out where the inital velocity comes into play, although it has to come in somewhere!

I believe you used the right kinematic equation in one of your other posts: it is $\Delta y = v_o t + \frac{1}{2}at^2$. If the initial velocity is zero, this reduces to $t^2 = 2 \Delta y/a$ as you have been using. In this question, use the general version.

 

[Medgirl314]

Oh! I'm not sure what LHS is, but it was most likely an odd accident. Thanks!


Thank you! Now I get it!

Δy=[v][0]t+1/2 a[t][2]
Δy=11.9 m/s(t) +1/2 [1.6(t)][2]

So what's next, since the inital velocity ISN'T zero? Do I divide both sides by t?

 

[Medgirl314]

Oh, I forgot I have Δy.

It should look like this:

44 m=11.9 m/s(t) +1/2 [1.6(t)][2]

So now I have everything but the time, so all I really have to do is simplify?

Thanks again!

 

[CAF123]

Oh, I forgot I have Δy.

It should look like this:

44 m=11.9 m/s(t) +1/2 [1.6(t)][2]

So now I have everything but the time, so all I really have to do is simplify?

Thanks again!

I would say so, but using a different kinematic eqn I obtain a different t, which is puzzling me. Perhaps another user could clarify.

 

[Medgirl314]

Hmm. That's odd. I wish I could help clarify, but I only know a few basic equations at this time. Do you think it would help if I tried attaching my formula sheet?

 

[CAF123]

Hmm. That's odd. I wish I could help clarify, but I only know a few basic equations at this time. Do you think it would help if I tried attaching my formula sheet?

When you solve your eqn, you will obtain two values of t since it is quadratic. If you average those values of t, you will obtain exactly the answer I get using another kinematic eqn. But I don't see the need to average and why there exists two times when the ball is at a height 44m. Let's wait for another user to clarify.

 

[Medgirl314]

Okay, thank you!

 

[Doc Al]

Oh, I forgot I have Δy.

It should look like this:

44 m=11.9 m/s(t) +1/2 [1.6(t)][2]

Careful with signs. The acceleration should be -1.6 m/s^2.

The real problem is that this equation is very sensitive to the exact initial velocity you use. Instead of using 11.9, use a more "exact" answer from your first part. Then the two solutions will converge to be the same.

Even better: Use distance = ave velocity * time. What's the average velocity?

 

[Medgirl314]

Okay, thanks!

vavg=Δx/Δt


Only I'm not sure of the average velocity. Should I use the previous equation and average my answers?

 

[Doc Al]

Only I'm not sure of the average velocity. Should I use the previous equation and average my answers?

No. For uniformly accelerated motion you can find the average velocity using this:
V_ave = (V₀ + V_f)/2

You already found the initial velocity earlier and you know the final velocity.

 

[Medgirl314]

Thanks! I can't believe I forgot that equation.

Something like this?

vavg= (11.9 m/s+0)/2≈ 5.95 m/s

 

[Doc Al]

Thanks! I can't believe I forgot that equation.

Something like this?

vavg= (11.9 m/s+0)/2≈ 5.95 m/s

Exactly. Now you can solve for the time to rise without worrying about any quadratics.

I just realized that they ask for the total time, not just the time to rise. Be careful!

You can use a quadratic equation to solve for the total time: It's easy!

 

[Medgirl314]

Okay, thank you! Do I use this equation next? vavg=Δx/Δt

 

[CAF123]

Just for completeness, to find the time taken to rise in one equation and without any quadratics consider using v = v₀ + at.

 

[Medgirl314]

Just to clarify, is this a formula I need to apply to this problem, or are you giving me a helpful formula for future reference? Thank you!

 

[CAF123]

Just to clarify, is this a formula I need to apply to this problem, or are you giving me a helpful formula for future reference? Thank you!

It is just another of the standard kinematic equations. I suggested it is an alternate way to find the time taken for the rock to rise to its apex. You know v, v₀ and g so you can find t - try it and see that it is the same as you get using the previous methods.

 

[Medgirl314]

Thank you! I will definitely keep this in mind. I don't have time to try it right now, because I am working on another three other physics problems, but it will prove useful very soon. Do you know the next step?

 

[CAF123]

Thank you! I will definitely keep this in mind. I don't have time to try it right now, because I am working on another three other physics problems, but it will prove useful very soon. Do you know the next step?

There are multiple ways to find the time taken for the rock to fall back down: Two are;

What is the rock's velocity at the top of the throw? (Let this be v₀)
How far does it fall? (Let this be Δy)
Now using one of the kinematic equations you can find v. Then use the kinematic equation I suggested in my last post to find t. (Be careful with signs)

or

Have you covered conservation of energy yet?

 

[Medgirl314]

Thank you! Are we using v=v0+at to find v?

 

[CAF123]

Thank you! Are we using v=v0+at to find v?

Provided you have all the correct data, you can use whatever equation you like - in this case, to find v from that eqn you would need to know the time taken for the rock to fall to the ground, but this is what you are looking for. Find instead a kinematic equation relating v₀, Δy and v so you can find v.

Once you know v, then use v = v₀+at to find t.

The time taken for the rock to fall back down is the same time it would take if somebody just dropped it from this height. This is because the rock's velocity is zero at the apex.

 

[Medgirl314]

The only other equation I can think of at the moment is v^2=v0^2+2aΔx, but that doesn't seem right somehow.

 

[CAF123]

The only other equation I can think of at the moment is v^2=v0^2+2aΔx, but that doesn't seem right somehow.

It is right, it relates v, v₀ and Δx. You know two of them so you can find the third.

 

[Medgirl314]

Great, thank you! I will do both and post my answer soon. Thanks for all the help!

 

[Medgirl314]

Sorry this reply took so long. Am I using 5.95 m/s or 0 for v? Thanks!

 

[haruspex]

Sorry this reply took so long. Am I using 5.95 m/s or 0 for v? Thanks!

In v_f² = v_i² + 2aΔx, v_i is the initial velocity and v_f is the final velocity. If you multiply all terms by a mass/2, you get the conservation of mechanical energy equation.

 

[Medgirl314]

Hi haurspex,

Thanks for the tip! unfortunately, I haven't quite gotten into mass yet, that is covered in my next chapter. Since it's a homework assignment, I'd prefer not to experiment, but I'll refer to that tip next chapter! I was wondering which one to use because 5.95 m/s is the average velocity. 0 m/s is the final velocity, but since someone else suggested we calculate the average, I wasn't sure which to use. Any ideas? Thanks again!

 

[haruspex]

I was wondering which one to use because 5.95 m/s is the average velocity. 0 m/s is the final velocity, but since someone else suggested we calculate the average, I wasn't sure which to use.

The reference to average velocity was distance = v_avg * time. You asked the question in the context of v²=v₀²+2aΔx, for which v is final velocity.
We can rewrite that using an average: v_avg * Δv = aΔx. (Because Δv = v - v₀ and v_avg = (v + v₀ )/2.)

 

[Medgirl314]

Thanks! Like this?
5.96 m/s^2 (5.95)=1.6 m/s^2 * 44 m

 

[haruspex]

Thanks! Like this?
5.96 m/s^2 (5.95)=1.6 m/s^2 * 44 m

Δv in this case would be 11.9m/s, no?

 

[Medgirl314]

Right! Thanks.

5.95 m/s^2 (11.9m/s^2)=1.6 m/s^2 * 44 m

 

[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?