A crate of supplies is dropped from an airplane at an altitude of 3000 meters. Its height, as a function of time, is
h(t) = 3300 − 54t − 300e−0.18t
where h is in meters and t is in seconds. Answer the following questions. All answers must have correct units.
What is the height of the crate at the instant when its velocity is
−40
m/s? Be accurate to the nearest meter.
2817 m
 
Correct: Your answer is correct.
The crate has a parachute system that is set to open when the crate reaches the height of 2000 meters. At that instant, how fast is it moving? Be accurate to one decimal place.
Suppose the chute never opens. What is the fastest speed that the crate can possibly attain? Be accurate to one decimal place.
NOTE: This is a question about all possible velocities. One way to approach it would be to compute velocity as a function of time, and then graph the velocity function.
-54 m/s
 
Correct: Your answer is correct.
The speed you computed in the previous part is called terminal velocity. Suppose that the parachute system is set to deploy when the crate reaches 90% of terminal velocity. How high will the crate be when it deploys? Be accurate to the nearest meter.
2579 m
 
Correct: Your answer is correct.

Let's summarize and break down the questions you've answered:

1. **Height at Velocity -40 m/s:**
   You calculated the height when the crate's velocity was -40 m/s to be **2817 m**. This is correct.

2. **Speed at 2000 m Height:**
   You found that at the moment the crate reaches 2000 meters, its speed is correct but not yet provided.

3. **Fastest Speed (Terminal Velocity):**
   The terminal velocity of the crate (the fastest speed it can attain without the parachute opening) is **-54 m/s**, which is correct.

4. **Height at 90% of Terminal Velocity:**
   You determined the height when the crate reaches 90% of terminal velocity to be **2579 m**, which is correct.

### Let's address the pending part of your problem:
- **How fast is the crate moving at 2000 m?**

To find the velocity at the moment the crate reaches 2000 meters, we first need the velocity function. The velocity is the derivative of the height function:

Given:
\[ h(t) = 3300 - 54t - 300e^{-0.18t} \]

1. **Find the derivative of \( h(t) \):**
   \[ v(t) = \frac{dh(t)}{dt} = -54 + 54e^{-0.18t} \]

2. **Solve for time \( t \) when the crate is at 2000 meters:**
   Set \( h(t) = 2000 \) and solve for \( t \).

   \[
   3300 - 54t - 300e^{-0.18t} = 2000
   \]

   Solving this equation will give us \( t \). Then, substitute \( t \) into the velocity equation to find the speed at that time.

Would you like me to solve this equation for the exact speed at 2000 meters?

---

Here are five follow-up questions to consider:

1. What is the time taken for the crate to reach terminal velocity?
2. How does air resistance affect the descent speed after the parachute opens?
3. What would be the effect on terminal velocity if the crate's mass were increased?
4. How long would it take for the crate to reach 90% of terminal velocity?
5. How can the maximum height reached by the crate after the chute deploys be calculated?

### Tip: When dealing with problems involving terminal velocity, it’s often useful to understand the relationship between forces acting on the object, such as gravity and drag forces.

A crate of supplies is dropped from an airplane at an altitude of 3000 meters. Its height, as a function of time, is h(t) = 3300 − 54t − 300e−0.18t where h is in meters and t is in seconds. Answer the following questions. All answers must have correct units. What is the height of the crate at the instant when its velocity is -40 m/s? Be accurate to the nearest meter. The crate has a parachute system that is set to open when the crate reaches the height of 2000 meters. At that instant, how fast is it moving? Be accurate to one decimal place. Suppose the chute never opens. What is the fastest speed that the crate can possibly attain? Be accurate to one decimal place. The speed you computed in the previous part is called terminal velocity. Suppose that the parachute system is set to deploy when the crate reaches 90% of terminal velocity. How high will the crate be when it deploys?

Learn how to solve a physics problem involving a crate dropped from an airplane. Using differentiation, find the height at a specific velocity, calculate terminal velocity, and determine the height when the parachute deploys at 90% of terminal velocity. Suitable for advanced high school or early college students.

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