Let's solve this step-by-step.

Part (a) - Finding the drag coefficient $b$

We are given that the diamond reaches a terminal velocity of 2.7 m/s. At terminal velocity, the frictional (drag) force balances the gravitational force. The frictional force is modeled by the equation:

$f = -bv$

At terminal velocity, $f = -bv_{\text{terminal}}$. The gravitational force is:

$f_{\text{gravity}} = mg$

At terminal velocity, these two forces are equal in magnitude:

$mg = bv_{\text{terminal}}$

Where:

• $m = 21.5 \, \text{g} = 0.0215 \, \text{kg}$ (converted to kilograms),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $v_{\text{terminal}} = 2.7 \, \text{m/s}$.

Now, solving for $b$:

$b = \frac{mg}{v_{\text{terminal}}}$

Substitute the values:

$b = \frac{(0.0215 \, \text{kg}) \times (9.8 \, \text{m/s}^2)}{2.7 \, \text{m/s}}$

$b = \frac{0.2107 \, \text{N}}{2.7 \, \text{m/s}} = 0.0780 \, \text{kg/s}$

So, $b = 0.0780 \, \text{kg/s}$.

Part (b) - Distance to reach 90% of terminal velocity

Next, we want to determine how far the diamond falls before it reaches 90% of its terminal velocity. The velocity of an object falling with air resistance follows the equation:

$v(t) = v_{\text{terminal}} \left( 1 - e^{-\frac{bt}{m}} \right)$

We want to find the time when the velocity reaches 90% of the terminal velocity:

$0.9 v_{\text{terminal}} = v_{\text{terminal}} \left( 1 - e^{-\frac{bt}{m}} \right)$

Simplifying:

$0.9 = 1 - e^{-\frac{bt}{m}}$

$e^{-\frac{bt}{m}} = 0.1$

Taking the natural logarithm of both sides:

$-\frac{bt}{m} = \ln(0.1)$

$t = -\frac{m}{b} \ln(0.1)$

Substitute the values for $m$ and $b$:

$t = -\frac{0.0215 \, \text{kg}}{0.0780 \, \text{kg/s}} \ln(0.1)$

$t = \frac{0.0215}{0.0780} \times 2.3026 \approx 0.6348 \, \text{s}$

Now, to find how far the diamond falls in this time, we use the equation for displacement under terminal velocity:

$y(t) = v_{\text{terminal}} \left( t - \frac{m}{b} \left( 1 - e^{-\frac{bt}{m}} \right) \right)$

Substitute the known values:

$y(0.6348) = 2.7 \, \text{m/s} \left( 0.6348 - \frac{0.0215}{0.0780} \left( 1 - e^{-\frac{0.0780 \times 0.6348}{0.0215}} \right) \right)$

Simplifying the terms:

$y(0.6348) = 2.7 \left( 0.6348 - 0.2756 \left( 1 - e^{-2.3026} \right) \right)$

Since $e^{-2.3026} \approx 0.1$:

$y(0.6348) = 2.7 \left( 0.6348 - 0.2756 \times 0.9 \right)$

$y(0.6348) = 2.7 \times (0.6348 - 0.2480)$

$y(0.6348) = 2.7 \times 0.3868 \approx 1.045 \, \text{m}$

So, the diamond falls approximately 1.045 meters before reaching 90% of its terminal velocity.

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Do you need further details or clarifications? Here are some related questions:

1. What factors affect the terminal velocity of an object?
2. How would the result change if the mass of the object was doubled?
3. Can this model apply to objects falling in air instead of water?
4. What is the difference between terminal velocity and free-fall acceleration?
5. How does the drag coefficient $b$ relate to the medium through which the object falls?

Tip: Terminal velocity depends on both the object's mass and the drag force exerted by the surrounding medium.