Stone Descent Calculation
The question refers to a stone being released from a balloon that is ascending at a rate of 5.0 m/s when the balloon is at a height of 535 m above the ground. We need to determine the time it takes for the stone to reach the ground.
First, we acknowledge that when the stone is released, it has the same upward velocity as the balloon, which is 5.0 m/s upwards. However, gravity will begin to act on the stone immediately after it is released, causing the stone to decelerate at a rate of approximately 9.81 m/s² (the acceleration due to gravity on Earth).
We can use the kinematic equation for uniformly accelerated motion to find the time it takes for the stone to reach the ground. The equation is given by:
\[ s = ut + \frac{1}{2}at^2 \]
where:
- \( s \) is the displacement (535 m in this case, but since the stone will travel upward before reversing direction, we will need to consider this in our calculations),
- \( u \) is the initial velocity (5.0 m/s upward),
- \( a \) is the acceleration due to gravity (-9.81 m/s², negative because it acts downward),
- \( t \) is the time in seconds.
However, we first need to calculate how much time it takes for the stone to come to a stop before it starts to fall back to the ground due to gravity. We use the following kinematic equation for that:
\[ v = u + at \]
Setting \( v = 0 \) since the stone comes to a stop momentarily when the initial upward speed is overcome by gravity, we get:
\[ 0 = 5.0 m/s - (9.81 m/s² \times t) \]
Now, solving for \( t \), which is the time for the stone to reach the peak of its trajectory:
\[ t = \frac{5.0 m/s}{9.81 m/s²} \approx 0.51 s \]
During this time, the stone will have traveled further upward. We calculate the upward distance using the first equation we mentioned:
\[ s = ut + \frac{1}{2}at^2 \]
\[ s = (5.0 m/s \times 0.51 s) + \frac{1}{2}(-9.81 m/s² \times (0.51 s)^2) \]
Plugging in the numbers we have:
\[ s \approx 5.0 m/s \times 0.51 s + \frac{1}{2} \times -9.81 m/s² \times 0.26 s^2 \]
\[ s \approx 2.55 m - 1.27 m \]
\[ s \approx 1.28 m \]
The total height from which the stone will now fall to the ground is the height of the balloon plus this additional ascent:
\[ H = 535 m + 1.28 m \]
\[ H \approx 536.28 m \]
Now, using the initial velocity as 0 m/s (since we are considering the point from which the stone starts falling down), and the total height from which it falls as \( H \), the kinematic equation becomes:
\[ 536.28 m = 0 m/s \times t + \frac{1}{2}(-9.81 m/s²)t^2 \]
This simplifies to:
\[ 536.28 m = -4.905 t^2 \]
Solving for \( t^2 \):
\[ t^2 = \frac{536.28 m}{4.905} \approx 109.3 \]
Now, solving for \( t \) (the time it takes for the stone to fall):
\[ t \approx \sqrt{109.3} \]
\[ t \approx 10.46 s \]
Finally, we sum the time it took for the stone to stop its ascent (0.51 s) and the time it took to fall to the ground (10.46 s):
Total time \( T = 0.51 s + 10.46 s \approx 10.97 s \)
Therefore, it will take approximately 10.97 seconds for the stone to reach the ground.
