To solve the problem of the two stones released from a rising balloon, we need to break it down into manageable parts. The balloon is ascending with a constant acceleration, and we have two stones released at different times. Let's analyze the motion of each stone and find the distance between them two seconds after the second stone is released.
Understanding the Motion of the Balloon
The balloon rises with a constant acceleration of 3 m/s². This means that its velocity increases by 3 m/s every second. We can use the equations of motion to describe the position of the balloon at any time.
Position of the Balloon
Let’s denote the time when the first stone is released as \( t = 0 \). The position of the balloon at time \( t \) can be expressed as:
Position of Balloon: \( s_b(t) = \frac{1}{2} a t^2 \)
Here, \( a = 3 \, \text{m/s}^2 \). So, the position of the balloon at any time \( t \) is:
\( s_b(t) = \frac{1}{2} \times 3 \times t^2 = 1.5 t^2 \)
Analyzing the Stones' Motion
Now, let's look at the two stones. The first stone is released at \( t = 0 \) seconds, and the second stone is released at \( t = 6 \) seconds. Both stones will experience free fall after being released, but they will also have an initial upward velocity equal to the velocity of the balloon at the moment of their release.
Velocity of the Balloon at Release Times
The velocity of the balloon at the time of release can be calculated using:
Velocity of Balloon: \( v_b(t) = a t \)
For the first stone (at \( t = 0 \)): \( v_b(0) = 3 \times 0 = 0 \, \text{m/s} \)
For the second stone (at \( t = 6 \)): \( v_b(6) = 3 \times 6 = 18 \, \text{m/s} \)
Position of Each Stone
Now we can find the position of each stone after their respective release times.
- First Stone: Released at \( t = 0 \) seconds.
- Position after \( t \) seconds: \( s_1(t) = s_b(0) + v_b(0) \cdot t - \frac{1}{2} g t^2 \)
- Since \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \), we have:
- \( s_1(t) = 0 + 0 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2 = -4.905 t^2 \)
- Second Stone: Released at \( t = 6 \) seconds.
- Position after \( t \) seconds (where \( t \) is the time after the second stone is released):
- \( s_2(t) = s_b(6) + v_b(6) \cdot t - \frac{1}{2} g t^2 \)
- Calculating \( s_b(6) \): \( s_b(6) = 1.5 \cdot 6^2 = 54 \, \text{m} \)
- Thus, \( s_2(t) = 54 + 18t - 4.905t^2 \)
Finding the Distance Between the Stones
We want to find the distance between the two stones 2 seconds after the second stone is released, which means we need to evaluate \( s_1(2) \) and \( s_2(2) \).
- For the first stone (2 seconds after the second stone is released, which is at \( t = 8 \) seconds total):
- \( s_1(8) = -4.905 \cdot (8^2) = -4.905 \cdot 64 = -313.92 \, \text{m} \)
- For the second stone (2 seconds after its release):
- \( s_2(2) = 54 + 18 \cdot 2 - 4.905 \cdot (2^2) = 54 + 36 - 19.62 = 70.38 \, \text{m} \)
Calculating the Distance Between the Stones
The distance between the two stones at that moment is:
Distance: \( |s_2(2) - s_1(8)| = |70.38 - (-313.92)| = 70.38 + 313.92 = 384.3 \, \text{m} \)
Therefore, the distance between the two stones 2 seconds after the second stone is released is approximately 384.3 meters. This illustrates how the initial conditions and the effects of gravity can lead to significant differences in the positions of objects in motion.
