# What is the acceleration of a falling object from a balloon?

1. Feb 21, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?"

2. Relevant equations
$v_B=12\frac{m}{s}$
$v_B(0)=0\frac{m}{s}$
$x_B=80m$
$x_B(0)=0m$
$v^2=v_0^2+2a(x-x_0)$
$x-x_0=\frac{1}{2}(v_0+v)t$
Answer to (a): 5.4 s; (b): 41 m/s

3. The attempt at a solution
$144\frac{m^2}{s^2}=2a(80m)$
$a_B=0.9{m}{s^2}$
$80m=\frac{1}{2}(12m\frac{m}{s})t_B$
$t_B=13.3s$
$a_P=9.8{m}{s^2}-0.9{m}{s^2}=8.9{m}{s^2}$
$-v_0^2=2(8.9\frac{m}{s^2})(0m-80m)$
$v_0^2=1424\frac{m^2}{s^2}$
$v_0=37.74\frac{m}{s}$
$-80m=\frac{1}{2}(-37.74\frac{m}{s})t$
$t=4.24s≠5.4s$

I can figure out the time of ascension and the acceleration of the balloon, but I don't know how to relate that to the package that it drops. Oddly, when I added the balloon's acceleration to $9.8m/s^2$, I came out with the correct velocity, but I can't seem to do the same for the time of descension for the package.

Last edited: Feb 21, 2016
2. Feb 21, 2016

### SteamKing

Staff Emeritus
What acceleration? The balloon is rising at presumably a constant velocity of 12 m/s. There's no indication that this velocity is changing, according to the problem statement.

You seem to be fixated on using only one of the SUVAT equations. Have you tried to see if one of the other SUVAT equations might be more applicable to this problem?

3. Feb 21, 2016

### Eclair_de_XII

Okay, so the acceleration of the balloon is zero. That doesn't affect the descent of the package? Anyway, my new variables for the package:

$a=9.8\frac{m}{s^2}$
$x=0m$
$x_0=80m$
$v=0m/s$

Find: $t$ and $v_0$. The only equation I can solve for, with only one unknown, is $x-x_0=vt-\frac{1}{2}at^2$, which gives me 4.04 s.

4. Feb 21, 2016

### SteamKing

Staff Emeritus
Remember, the balloon is ascending at 12 m/s when the package is dropped over the side. What does that tell you about the motion of the package initially?

5. Feb 21, 2016

### Eclair_de_XII

It's initially 12 m/s? So if it's falling, it would be negative?

6. Feb 21, 2016

### SteamKing

Staff Emeritus
Is the package falling initially when it first leaves the balloon? The balloon's going up. What does Mr. Newton say about objects which are already in motion?

7. Feb 21, 2016

### Eclair_de_XII

They stay in motion?

8. Feb 21, 2016

### SteamKing

Staff Emeritus
Correct, unless acted on by another force. Which force acts on the package once it's left the balloon?

9. Feb 21, 2016

### Eclair_de_XII

Gravity!

10. Feb 21, 2016

### SteamKing

Staff Emeritus
And what does gravity first do to the package before it can fall to the ground?

11. Feb 21, 2016

### Eclair_de_XII

Accelerates it down at $9.8 m/s^2$?

12. Feb 21, 2016

### SteamKing

Staff Emeritus
Yes, that happens. But what occurs to the initial velocity which the package had as it first left the balloon?

13. Feb 21, 2016

### Eclair_de_XII

I suppose it drops to zero?

14. Feb 21, 2016

### SteamKing

Staff Emeritus
And how long would that take?

15. Feb 21, 2016

### Eclair_de_XII

Let's see if I got this right:

$v=0\frac{m}{s}$
$v_0=12\frac{m}{s}$
$a=-9.8\frac{m}{s^2}$

$v=v_0+at$
$0\frac{m}{s}=12\frac{m}{s}-(t)9.8\frac{m}{s^2}$
$-12\frac{m}{s}=-(t)9.8\frac{m}{s^2}$
$t=1.2245 s$

Although I'm not sure on why acceleration should be negative. It's because it's going to cancel out the 12 m/s, right?

16. Feb 21, 2016

### SteamKing

Staff Emeritus
These questions are usually decided when setting the problem up initially. A common assumption is that up is positive and down is negative.

Now the package was 80 m above the ground when it left the balloon. Is it still 80 m above the ground or has it moved? How high is the package when its upward velocity drops to zero?

17. Feb 21, 2016

### Eclair_de_XII

$v=0\frac{m}{s}$
$v_0=12\frac{m}{s}$
$x_0=80m$
$t=1.2245s$

$x-x_0=\frac{1}{2}(v_0+v)t$
$x=80m+\frac{1}{2}(12\frac{m}{s})(1.2245s)$
$x=87.347m$

I know for a fact that the distance between the package and the ground cannot be greater than eighty meters. How would I go about reasoning that it's actually: $x=80m-\frac{1}{2}(12\frac{m}{s})(1.2245s)=72.653m$ and not plus?

18. Feb 21, 2016

### SteamKing

Staff Emeritus
1. The package is 80 m above the ground when it initially leaves the balloon.

2. We've already established the package has an initial upward velocity of 12 m/s, the same as the balloon.

3. Mr. Newton says the package remains in its original motion, which is upward, after leaving the balloon.

4. Yet you say the package cannot be more than 80 meters above the ground, even after you have calculated that it takes 1.22 s for the velocity of the package to drop to zero, so that it can begin to free fall back to earth.

Interesting.

I guess all that guff about Newton being a great physicist was wrong.

19. Feb 21, 2016

### Eclair_de_XII

So... I guess 87.347 m is right...?

20. Feb 21, 2016

### Eclair_de_XII

Oh, so during that 1.22 s, it travels upward 7.35 m?