A stone is dropped from an ascending balloon...

[tremain74]

TL;DR Summary: I have a problem that I am working on in my Engineering Dynamics Book. A stone is dropped from a balloon that is ascending at a uniform rate of 10m/s. If it takes the stone 10 seconds to reach the ground, how high was the balloon at the instant the stone was dropped? The answer is 485 meters.

Based on the information, I believe that uniform rate would mean no acceleration. Therefore I used the formula s = 1/2 (v+v0)*t. Therefore my formula was s = 1/2(10+0)*10 where initial velocity is 0 and final velocity is 10. Also it would take ten seconds to reach the ground but my answer was wrong.

 

[.Scott]

I don't get 485 meters.
Let's estimate the acceleration from gravity at 10m/ss.
You seem to be claiming that the initial velocity of the stone is 0m/s and it strikes the ground at 10m/s. Both of those are wrong.

Let's start with you answering two questions: 1) What is the initial velocity of the stone? That it, at the moment it detaches from the balloon. 2) Once the stone is falling, how much will its velocity change in 10 seconds?

 

[.Scott]

@tremain74 : I would like to work you through the problem. But you seem to have left.

 

[tremain74]

I am back. I used the formula s = 1/2(v+v0)*t. The first was s0 = 1/2*(10+0)*1. The answer was 5. My logic was at first the stone left the balloon starting the velocity as 0. Then I continued to use the formula s1 =1/2 (v+v0)*2 for two seconds and ect. every time I used the formula I add a second to the time that the stone was ascending towards the ground. Also, I thought that uniform rate was no acceleration.

 

[tremain74]

I am back. I used the formula s = 1/2(v+v0)*t. The first was s0 = 1/2*(10+0)*1. The answer was 5. My logic was at first the stone left the balloon starting the velocity as 0. Then I continued to use the formula s1 =1/2 (v+v0)*2 for two seconds and ect. every time I used the formula I add a second to the that the stone was ascending towards the ground. Also, I thought that uniform rate was no acceleration.

 

[.Scott]

Hmmm. It appears that you have been stuck on this problem for 14 months. You asked a remarkably similar question on March 20, 2024. At that time, @Babadag came up with a way of interpreting the question that yielded about 485 meters. He used acceleration due to gravity at 9.8m/ss, he took the 10 second time as only the path from the stones peak altitude (well after it detached from the balloon) to when it hit the ground, but he took the "drop" time to be when the stone detached from the balloon. That's certainly a creative way of getting 485 meters.

 

[.Scott]

The initial velocity is not zero. At the moment that the stone detached from the balloon, would it not be traveling at the same speed as the balloon?
The uniform rate is the velocity of the balloon and the stone as they rise together. Once the stone is released, it will free-fall due to gravity. And accelerate due to gravity.

 

[tremain74]

I actually forgot that I worked on this problem from last year. But I now remember.

 

[.Scott]

So let's put this to bed this time.
What is the initial velocity of the stone - at the moment that it detaches from the balloon.

And forget about that 485 meter answer.

 

[tremain74]

It's 10 meters per second

 

[.Scott]

I can't stick around. So I will walk you through it:
What is the velocity of the stone when it is released?
A:

Spoiler

10m/s in the upward direction. I'm going to call up negative, so -10m/s.

Over the 10 seconds before it strike the ground, how much will it velocity change?
A:

Spoiler

Using 10m/ss for gravity, 10 seconds time 10m/ss = 100m/s

So what will be the final velocity?
A:

Spoiler

(-10 + 100)m/s = 90m/s

Since it is accelerating smoothly and linearly, the average velocity will be the average at start and end.
A:

Spoiler

start=-10, end=90, average = (-10+90)/2 m/s = 40m/s.

So what will be the total drop in altitude during those 10 seconds:
A:

Spoiler

10 second X 40m/s = 400 meters

There is a formula: $s = (at^2)/2 + v_0 t$
I will let others work this with you.
Have fun and see you next year.