What is the velocity​ (in 

ft divided by sft/s​) 

of a sandbag 

2.752.75 

s after it is released from a​ hot-air balloon that is stationary in the​ air?

To find the velocity of the sandbag 2.75 seconds after it is released from a stationary hot-air balloon, we can use the same approach as before:

### Given:
- Initial velocity (\( v_0 \)) = 0 ft/s (because the balloon is stationary)
- Acceleration due to gravity (\( g \)) = 32 ft/s²
- Time (\( t \)) = 2.75 s

### Formula:
\[
v(t) = v_0 + g \cdot t
\]

Since \( v_0 = 0 \), the equation simplifies to:

\[
v(t) = g \cdot t
\]

### Substituting the values:
\[
v(2.75) = 32 \, \text{ft/s}^2 \times 2.75 \, \text{s}
\]

\[
v(2.75) = 88 \, \text{ft/s}
\]

### Answer:
The velocity of the sandbag 2.75 seconds after it is released is \( 88 \, \text{ft/s} \) downward.

Would you like more details or have any questions?

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Here are some related questions you might consider:
1. What is the height from which the sandbag was released if it hits the ground after 5 seconds?
2. How would the velocity change if the balloon was ascending at a constant speed when the sandbag was released?
3. What would be the velocity after 5 seconds of free fall?
4. How much time would it take for the sandbag to reach a velocity of 160 ft/s?
5. How does the mass of the sandbag affect its velocity during free fall (assuming no air resistance)?

**Tip:** When calculating velocity in free-fall problems, the time and acceleration due to gravity are the key factors, assuming no air resistance.

Learn how to calculate the velocity of a sandbag released from a stationary hot-air balloon after 2.75 seconds of free fall. This problem involves kinematics and the acceleration due to gravity formula. Suitable for advanced high school students.

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