To solve the problem of the balloon rising and the stone being dropped, we need to analyze the motion of both the balloon and the stone. The key here is to understand the relationship between the time it takes for the stone to reach the ground and the acceleration of the balloon.
Understanding the Motion of the Stone
When the stone is dropped from the balloon, it begins to fall under the influence of gravity. The time it takes for the stone to hit the ground is given as \( \frac{2k}{g} \), where \( g \) is the acceleration due to gravity. This time is independent of the height from which the stone is dropped, which suggests that the balloon is accelerating upwards at a constant rate.
Analyzing the Stone's Fall
Using the equations of motion, we can express the distance \( h \) that the stone falls in terms of time \( t \). The equation for the distance fallen under constant acceleration due to gravity is:
- \( h = \frac{1}{2} g t^2 \)
However, since the stone is dropped from a balloon that is also moving upwards, we need to account for the initial height of the balloon and the upward motion of the balloon during the time \( t \). If we denote the initial height of the balloon at the moment the stone is dropped as \( H \), the equation becomes:
- \( H - h = H - \frac{1}{2} g t^2 \)
Relating Time and Balloon's Acceleration
Given that the time \( t \) for the stone to reach the ground is \( \frac{2k}{g} \), we can substitute this into our equation. The height \( H \) can also be expressed in terms of the balloon's acceleration \( a \) and the time \( t \) it has been rising:
- \( H = \frac{1}{2} a t^2 \)
Substituting \( t = \frac{2k}{g} \) into this equation gives:
- \( H = \frac{1}{2} a \left(\frac{2k}{g}\right)^2 = \frac{2k^2 a}{g^2} \)
Setting Up the Equation
Now we can equate the two expressions for \( H \):
- \( \frac{2k^2 a}{g^2} = \frac{1}{2} g \left(\frac{2k}{g}\right)^2 \)
On simplifying the right side:
- \( \frac{1}{2} g \cdot \frac{4k^2}{g^2} = \frac{2k^2}{g} \)
Now we have:
- \( \frac{2k^2 a}{g^2} = \frac{2k^2}{g} \)
Finding the Acceleration
By canceling \( 2k^2 \) from both sides (assuming \( k \neq 0 \)), we arrive at:
- \( \frac{a}{g^2} = \frac{1}{g} \)
Multiplying both sides by \( g^2 \) gives:
Conclusion on Balloon's Acceleration
The acceleration of the balloon, therefore, is constant and equal to the acceleration due to gravity, \( g \). This means that as the balloon rises, it does so with an upward acceleration equal to the gravitational pull acting downwards. This result is quite interesting as it shows that the balloon's acceleration is not influenced by the height from which the stone is dropped, reinforcing the idea of uniform acceleration in this scenario.
