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

[Eclair_de_XII]

Homework Statement

"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?"

Homework 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

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.

 

[SteamKing]

Homework Statement

"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 odes the package take to reach the ground? (b) With what speed does it hit the ground?"

Homework 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

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.

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?

 

[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.

 

[SteamKing]

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.

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?

 

[Eclair_de_XII]

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

 

[SteamKing]

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

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?

 

[Eclair_de_XII]

They stay in motion?

 

[SteamKing]

They stay in motion?

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

 

[Eclair_de_XII]

Gravity!

 

[SteamKing]

Gravity!

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

 

[Eclair_de_XII]

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

 

[SteamKing]

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

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

 

[Eclair_de_XII]

I suppose it drops to zero?

 

[SteamKing]

I suppose it drops to zero?

And how long would that take?

 

[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?

 

[SteamKing]

Let's see if I got this right:

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

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

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?

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?

 

[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?

 

[SteamKing]

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

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

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?

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.

 

[Eclair_de_XII]

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

 

[Eclair_de_XII]

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.

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

 

[SteamKing]

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

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

Yes and Yes.

Now you can calculate how long it takes the package to fall from this height.

Remember, part a) asks how long it takes for the package to hit the ground (after it leaves the balloon.)

You should also be able to answer part b) without any problem now that you know from what height the package free falls.

 

[Eclair_de_XII]

$-v_0^2=2(9.8\frac{m}{s^2})(-87.347m)$
$v_0^2=1712\frac{m^2}{s^2}$
$v_0=41.38\frac{m}{s}$

$-41.38\frac{m}{s}=(-9.8\frac{m}{s^2})t_1$
$t_1=4.222s$
$t+t_1=5.4\frac{m}{s}$

Funny thing is, when I try to use the $x-x_0=v_0t+\frac{1}{2}at^2$ formula, I either have to switch the values for $x_0$ and $x$, or change the sign of gravitational acceleration.

 

[Eclair_de_XII]

One more question: if this balloon was moving horizontally, would I have to take the sine of the velocity of the balloon and reduce that to zero with gravitational acceleration?

Or maybe reduce the y-part of the balloon's velocity vector?

 

[SteamKing]

One more question: if this balloon was moving horizontally, would I have to take the sine of the velocity of the balloon and reduce that to zero with gravitational acceleration?

Or maybe reduce the y-part of the balloon's velocity vector?

Remember, gravity acts in only one direction: it pulls objects toward the center of the earth. If the balloon is still moving upward at 12 m/s, the horizontal velocity component will not affect the time it takes for the package to reach the ground. The point at which it hits, relative to where is was released, will be different.

 

[Eclair_de_XII]

So you would reduce the whole vector, and not its y-component?

 

[SteamKing]

So you would reduce the whole vector, and not its y-component?

Re-read my post again very carefully. Gravity acts in only one direction, toward the center of the earth. Gravity cannot arrest horizontal motion.

 

[Eclair_de_XII]

Oh, and I think I'm supposed to change the acceleration value once the initial velocity is reduced to zero. When it's being reduced to zero, it's decelerating, which means that a is negative.

When it falls back down, it's accelerating, so it's positive?

 

[Eclair_de_XII]

Gravity cannot arrest horizontal motion.

Forgive me for being dense, but if gravity cannot affect horizontal motion, and a whole vector is affected by the x-component, then one would have to reduce only the y-component, to zero, right?

 

[SteamKing]

Forgive me for being dense, but if gravity cannot affect horizontal motion, and a whole vector is affected by the x-component, then one would have to reduce only the y-component, to zero, right?

Correct.

There's no law in mechanics which says that all components of a vector must be changed equally. Only those components actually affected by something like gravity will change. All others stay the same (unless there's a headwind, for example).

 

[SteamKing]

$-v_0^2=2(9.8\frac{m}{s^2})(-87.347m)$
$v_0^2=1712\frac{m^2}{s^2}$
$v_0=41.38\frac{m}{s}$

I think here, instead of introducing spurious negative signs, it would be better to use g = -9.8 m/s² along with the distance fallen of -87.347 m.
v₀² still comes out positive.

$-41.38\frac{m}{s}=(-9.8{m}{s^2})t_1$
$t_1=4.222s$
$t+t_1=5.4\frac{m}{s}$

Funny thing is, when I try to use the $x-x_0=v_0t+\frac{1}{2}at^2$ formula, I either have to switch the values for $x_0$ and $x$, or change the sign of gravitational acceleration.

See above. g will be negative since that's what we chose to indicate motion toward the ground, i.e. down. The change in position will also be negative since the package was 87.347 m above the ground before it fell.

 

[Eclair_de_XII]

I think here, instead of introducing spurious negative signs, it would be better to use g = -9.8 m/s2 along with the distance fallen of -87.347 m.
v02 still comes out positive.

But won't it be imaginary since $v_0$ is still negative?

The change in position will also be negative since the package was 87.347 m above the ground before it fell.

But then you'd have to move the negative displacement to the other side, making it positive.

 

[Eclair_de_XII]

Well, anyway, thank you so much for everything. You and everyone have been a big help to me with my self-study. Without this forum, I don't believe I would be able to learn physics on my own. So... thank you. :)

 

[SteamKing]

But won't it be imaginary since $v_0$ is still negative?

The equation is v² = u² + 2 a s

There is no negative sign on the velocity terms. The product of -9.8 and -87.347 is also positive.

But then you'd have to move the negative displacement to the other side, making it positive.

Not if you take x = 0 being the ground and x₀ = 87.347 m as the height from which the package started falling. Here, v₀ = 0 and g = -9.8 m/s², making the equation:

x - x₀ = v₀ t + (1/2)*g*t² → 0 - 87.347 = (1/2) * (-9.8) * t²