A hot-air balloon is rising straight upward with a constant speed of 6.50 m/s. When the basket of the
balloon is 20.0 m above the ground, a bag of sand tied to the basket comes loose.
- Find the speed of the bag just before it lands when it is released from 20.0 m.
- What is the amount of time the bag is in the air?

We can solve this problem using kinematic equations, as the motion of the bag follows the rules of uniformly accelerated motion under gravity.

### Given Data:
- Initial height of the basket (and the sandbag when released), \( h_0 = 20.0 \, \text{m} \)
- Initial velocity of the bag, \( v_0 = 6.50 \, \text{m/s} \) (since the balloon is rising upward)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Final height (just before the bag hits the ground), \( h_f = 0 \, \text{m} \)

### 1. Speed of the bag just before it lands:
We'll use the following kinematic equation that relates initial velocity, final velocity, displacement, and acceleration:

\[
v_f^2 = v_0^2 + 2g \cdot (h_f - h_0)
\]

Where:
- \( v_f \) is the final velocity (speed of the bag just before hitting the ground)
- \( v_0 = 6.50 \, \text{m/s} \) (initial upward velocity)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( h_f - h_0 = 0 - 20 = -20 \, \text{m} \) (since the bag is falling downward, this displacement is negative)

Plugging in the values:

\[
v_f^2 = (6.50)^2 + 2 \times 9.81 \times (-20)
\]
\[
v_f^2 = 42.25 - 392.4
\]
\[
v_f^2 = -350.15
\]

Taking the square root (and noting that the velocity will be downward, so we take the negative root):

\[
v_f = -\sqrt{350.15} = -18.71 \, \text{m/s}
\]

Thus, the **speed** of the bag just before it lands is \( 18.71 \, \text{m/s} \).

### 2. Time the bag is in the air:
We can use the following kinematic equation to find the time \( t \) the bag is in the air:

\[
h_f = h_0 + v_0 t - \frac{1}{2} g t^2
\]

Where:
- \( h_f = 0 \, \text{m} \)
- \( h_0 = 20.0 \, \text{m} \)
- \( v_0 = 6.50 \, \text{m/s} \)
- \( g = 9.81 \, \text{m/s}^2 \)

Substituting these values into the equation:

\[
0 = 20 + 6.50t - \frac{1}{2} \times 9.81 t^2
\]
\[
0 = 20 + 6.50t - 4.905t^2
\]

This is a quadratic equation:

\[
4.905t^2 - 6.50t - 20 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where:
- \( a = 4.905 \)
- \( b = -6.50 \)
- \( c = -20 \)

Substituting the values:

\[
t = \frac{-(-6.50) \pm \sqrt{(-6.50)^2 - 4(4.905)(-20)}}{2(4.905)}
\]
\[
t = \frac{6.50 \pm \sqrt{42.25 + 392.4}}{9.81}
\]
\[
t = \frac{6.50 \pm \sqrt{434.65}}{9.81}
\]
\[
t = \frac{6.50 \pm 20.85}{9.81}
\]

Taking the positive root (since time cannot be negative):

\[
t = \frac{6.50 + 20.85}{9.81} = \frac{27.35}{9.81} \approx 2.79 \, \text{seconds}
\]

Thus, the **time the bag is in the air** is approximately \( 2.79 \, \text{seconds} \).

---

### Summary of Results:
1. The speed of the bag just before it lands is \( 18.71 \, \text{m/s} \).
2. The time the bag is in the air is \( 2.79 \, \text{seconds} \).

Would you like more details or have any questions about these calculations?

Here are 5 related questions:
1. How would the problem change if the balloon was descending when the bag was released?
2. What if the balloon was rising with a different speed, say 10 m/s?
3. How can we calculate the maximum height the bag would reach if it was released while the balloon was rising faster?
4. How would we calculate the velocity of the bag at any specific time after release?
5. What effect would air resistance have on the motion of the bag?

**Tip:** When using the quadratic formula, always check whether both roots are valid based on the physical context of the problem.

A hot-air balloon is rising straight upward with a constant speed of 6.50 m/s. When the basket of the balloon is 20.0 m above the ground, a bag of sand tied to the basket comes loose. Find the speed of the bag just before it lands when it is released from 20.0 m. What is the amount of time the bag is in the air?

Solve a physics problem where a bag of sand is released from a rising hot-air balloon at 6.50 m/s from a height of 20.0 m. Learn how to calculate the bag's speed just before landing and the time it spends in the air using kinematic equations. Suitable for Grades 10-12.

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