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?

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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.