# Physics HW- Falling Bodies, Part 3

1. Nov 6, 2013

### Medgirl314

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
v0^2=v^2-2αΔy
t^2 =2y/a

3. 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!

2. Nov 6, 2013

### CAF123

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

3. Nov 6, 2013

### 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!!!

4. Nov 6, 2013

### CAF123

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

5. Nov 6, 2013

### 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!

6. Nov 6, 2013

### CAF123

vo2=v2-2aΔy is the same as v2=vo2+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.

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.

7. Nov 6, 2013

### 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?

8. Nov 7, 2013

### 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!

9. Nov 7, 2013

### CAF123

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

10. Nov 7, 2013

### 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?

11. Nov 7, 2013

### CAF123

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.

12. Nov 7, 2013

### Medgirl314

Okay, thank you!

13. Nov 7, 2013

### Staff: Mentor

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?

14. Nov 7, 2013

### 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?

15. Nov 8, 2013

### Staff: Mentor

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

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

16. Nov 8, 2013

### Medgirl314

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

Something like this?

vavg= (11.9 m/s+0)/2≈ 5.95 m/s

17. Nov 8, 2013

### Staff: Mentor

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!

18. Nov 8, 2013

### Medgirl314

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

19. Nov 8, 2013

### CAF123

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

20. Nov 8, 2013

### 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!