# Falling Object

1. Sep 16, 2007

### Heat

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

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground. After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
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What is the greatest height above the ground that the sandbag reaches?

3. The attempt at a solution

I would logically think that the greatest height above the ground that it reaches is 40m, as that's when it gets released.

The feedback I got this incorrect answer is to recheck calculation, or that I may have made rounding errors or sig figs.

and if I do it algebraically: it would look something like this ::

vx = vo + at

40.0 + (-9.8) (0)
= 40.0

2. Sep 16, 2007

### Staff: Mentor

If not for the fact that the balloon was moving, you'd be correct.
Not sure what you're doing here.

Hint: What's the initial vertical velocity of the sandbag?

3. Sep 16, 2007

### Heat

50m/s.
.......
"releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground"

.........

if the person releases it an the exact instant when the balloon is 40.0.

Meaning that he/she is not touching it anymore, and lets go of it, how could the bag still go up. I would go down, and making sense that the max height it achieved was when it was let go of.

4. Sep 16, 2007

### Staff: Mentor

What's this?
Because it has an initial upward velocity. (Remember that it was moving with the balloon when it was released.) As it goes up, gravity will slow it down until it reaches a maximum height, then starts to fall back down.

Just like when you throw a ball up into the air. When it leaves your hand it continues moving up, doesn't it?

5. Sep 16, 2007

### Heat

woudnt initial velocity be 50m/s as the ballon is being exerted up?

6. Sep 16, 2007

### Staff: Mentor

But the problem states:

7. Sep 16, 2007

### Heat

woops I misread the problem :( , you are right it is 5.00 m/s.

so I would solve using the kinematics equation..

time = 3.41
v = 28.4
final x = 0
initial x = ?
initial v = 5

x-x0 = t/2 (vo +vx)
x = 3.41/2 (5+28.4)
x = 56.947

8. Sep 16, 2007

### Staff: Mentor

Not sure what you're doing here. What is this time? What is v?

Try this: How long (how many seconds) before it reaches maximum height?

9. Sep 16, 2007

### Heat

the seconds it takes to reach ground is 3.41
and the velocity at which it reaches ground is 28.4

he should reach the height of 40meters in 8 seconds, if the info carries of velocity 5 m/supwards and reaches 40 meters.

10. Sep 16, 2007

### Staff: Mentor

Ah.... I see. Not sure why you calculated all that, but you could use it to find the answer.

Don't understand this at all. If it goes up to max height, then back down to the ground in 3.41 seconds, then it must reach max height sometime before 3.41 seconds. Right?

Again, try this: When it's dropped off the balloon, it's moving at + 5 m/s. How long does it take from that point to reach max height? (Hint: What is the speed at max height?)

11. Sep 17, 2007

### learningphysics

The moment the bag is released, it is still moving upwards. How far upwards does it travel... use a kinematics equation to find out...

12. Sep 17, 2007

### Heat

time in which bag is release is -3.41 (since it reaches ground at 3.41)
x0 40m upwards.
acceleration is 9.8m/s^2
xf=?

is this how I set it up?

13. Sep 17, 2007

### Staff: Mentor

There are several ways to solve this using kinematics; here are two:

(1) You can directly find out how high it will move from the initial position, using a kinematic formula relating velocity, distance, and acceleration. (I think this is what learningphysics had in mind. It's the quickest way.)

(2) You can find the time it takes to reach its max height, using the relationship between velocity, acceleration, and time. Then plug the time into your height equation. (This is a bit longer, but might be easier to understand.)

14. Sep 17, 2007

### Heat

Taking option one:
Velocity Final^2 = Velocity Initial ^2 + 2 ( Acceleration)(X Final - X Initial)

28.4^2 = 5 ^2 + 2 ( -9.8)(X Final - 0)
806.56 = 25 + -19.6X Final
781.56 = -19.6X Final
-39.88 = X Final

Although this is incorrect...

The velocity Final I am taking when it lands, although I think that is the part that is messing this up.

Velocity Initial is 5 because the the balloon is going up.

Last edited: Sep 17, 2007
15. Sep 17, 2007

### Staff: Mentor

Just find the distance traveled between the point where it is dropped (1) and the point where it reaches maximum height (2). Point 1 has: v = 5; h = 40. Point 2 has v = ?; h = ? (I want you to tell me what speed it has at its highest point.)

16. Sep 17, 2007

### Tom Winter

Inertia. Newton. The entire contents of the balloon, people and the sandbag alike, share the velocity of the overall vehicle. To negate the upward velocity of the balloon, the person would have to THROW it down, at a downward speed equal to the balloon's upward speed, rather than just drop it.

17. Sep 17, 2007

### Heat

v at max height has to be 0.

and thanks for clarifying that winter. My mind just opened up when I thought of the question the other way around.

18. Sep 17, 2007

### Staff: Mentor

Yes!

19. Sep 17, 2007

### Heat

after doing the problem I got 1.28

so I add that to 40 and I get 41.28.

Does this seems right?

20. Sep 17, 2007

### learningphysics

looks good to me.