A hot air balloon rises at a constant acceleration

In summary, the sandbag is released from a hot air balloon with a constant acceleration of 4 m/s^2 and rises to a height of 16 meters. The sandbag is then released and falls with an acceleration of -9.8 m/s^2 due to Earth's gravity. The time taken for the sandbag to hit the ground can be calculated using the equation s=So+(Vo)(t)+1/2(a)(t)^2 and plugging in the values of So=16 m, Vo=11.31 m/s, and a=-9.8 m/s^2. The resulting quadratic equation can then be solved to find the time taken for the sandbag to hit the ground, which is approximately
  • #1
Girn261

Homework Statement


A hot air balloon is released from the ground & rises up with a constant acceleration of 4 m/s^2. When it is 16 meters above a sandbag is dropped.

What is the time taken for the sandbag to hit the ground?

Homework Equations


v^2=u^2+2as
v=u+at

The Attempt at a Solution


v^2=u^2+2as
v^2=0+2(4)(16)
v=11.31m/s

v=u+at
11.31=0+(4)(t)
t=2.828
 
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  • #2
Would that be the time taken for the sandbag to hit the ground ?
Or is it the time the balloon plus bag take to rise 16 m and reach a speed of 11.3 m/s?
 
  • #3
You found the sandbag is traveling upwards w.r.t. the Earth when it is released at a height of 16 meters. After this, it will accelerate downward from Earth's gravity with ## a=-9.8 ## m/sec^2. You need to write the equation of its motion, ## s ## vs. ## t ## starting from the time it is released. The problem has a couple of steps to it, but this ## s ## vs. ## t ## equation is an important one.
 
  • #4
Charles Link said:
You found the sandbag is traveling upwards w.r.t. the Earth when it is released at a height of 16 meters. After this, it will accelerate downward from Earth's gravity with ## a=-9.8 ## m/sec^2. You need to write the equation of its motion, ## s ## vs. ## t ## starting from the time it is released. The problem has a couple of steps to it, but this ## s ## vs. ## t ## equation is an important one.

s=ut+1/2at^2
16=0+1/2*(9.81)(t)^2
t=1.806s

So I assume this is the time taken to reach the ground after the upward descent from the acceleration of the hot air balloon. But if I add the two together (1.806+2.828) I get the wrong answer.
 
  • #5
Girn261 said:
s=ut+1/2at^2
16=0+1/2*(9.81)(t)^2
t=1.806s

So I assume this is the time taken to reach the ground after the upward descent from the acceleration of the hot air balloon. But if I add the two together (1.806+2.828) I get the wrong answer.
Your equation in the form you have it is incorrect. The equation has the form ## s=s_o+v_o t+(1/2)a t^2 ##. You need to figure out what ## s_o ##, ## v_o ##, and ## a ## need to be. Also what is ## s ## when the sandbag hits the ground? In addition, the ## t ## in this formula is a clock that starts when the sandbag is released. That part I think you already figured out though. ## \\ ## And additional comment: For the equation of ## s ## vs. ## t ##, there are a couple of options available, e.g. you can choose up as either positive or negative, but you need to be consistent. There is also a choice of where to select the origin for ## s ##, i.e.the place where ## s=0 ##, but again, you need to be consistent.
 
Last edited:
  • #6
Here's my attempt
S=So+(Vo)(t)+1/2(a)(t)^2
0=16+0+(1/2(9.81+4)t^2)
T=-4.79s

I tell you, physics is one of my worst subjects lol. Ugh.. confusing figuring out what's initial and what's final, which acceleration to use. Etc
 
  • #7
Girn261 said:
Here's my attempt
S=So+(Vo)(t)+1/2(a)(t)^2
0=16+0+(1/2(9.81+4)t^2)
T=-4.79s

I tell you, physics is one of my worst subjects lol. Ugh.. confusing figuring out what's initial and what's final, which acceleration to use. Etc
The sandbag is moving (quickly) upward (w.r.t. the earth) as the rope is cut. Because of this, the ## v_o ## is not zero, but rather the number you previously computed. ## \\ ## Also the acceleration ## a ## after the rope is cut is incorrect. It is much simpler than that. The only force on the sandbag after the rope is cut is due to gravity=this should tell you what the acceleration will be.
 
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  • #8
Charles Link said:
The sandbag is moving (quickly) upward (w.r.t. the earth) as the rope is cut. Because of this, the ## v_o ## is not zero, but rather the number you previously computed. ## \\ ## Also the acceleration ## a ## after the rope is cut is incorrect. It is much simpler than that. The only force on the sandbag after the rope is cut is due to gravity=this should tell you what the acceleration will be.

Thanks for the help boss, I realized my mistake, I plugged the correct units into the formula. And then used the quadratic formula to get my answer.

Cheers
 
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Related to A hot air balloon rises at a constant acceleration

1. What causes a hot air balloon to rise at a constant acceleration?

The hot air balloon rises at a constant acceleration due to the principle of buoyancy. As the air inside the balloon is heated, it becomes less dense than the surrounding air and therefore experiences an upward force, causing the balloon to rise.

2. How is the acceleration of a hot air balloon determined?

The acceleration of a hot air balloon is determined by factors such as the size and weight of the balloon, the temperature difference between the air inside and outside the balloon, and the rate at which the balloon is heated. These factors can be calculated using mathematical equations.

3. Can the acceleration of a hot air balloon be controlled?

Yes, the acceleration of a hot air balloon can be controlled by adjusting the amount of heat in the balloon. Adding more heat will increase the acceleration and cause the balloon to rise faster, while reducing the heat will decrease the acceleration and cause the balloon to descend.

4. Does the acceleration of a hot air balloon change during the flight?

The acceleration of a hot air balloon may change during the flight as the temperature of the air inside the balloon and the surrounding air may vary. However, as long as the temperature difference is maintained, the balloon will continue to rise at a constant acceleration.

5. Are there any safety concerns regarding the constant acceleration of a hot air balloon?

There are some safety concerns regarding the constant acceleration of a hot air balloon, such as maintaining a safe distance from other balloons and monitoring the temperature and pressure inside the balloon to ensure it does not exceed safe limits. It is important for pilots to have proper training and follow safety protocols to ensure a safe flight.

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