Answer:
The drag force on an object is given by the equation:
F_d = -bv^2
where F_d is the drag force, b is a constant that depends on the properties of the fluid and the shape of the object, and v is the velocity of the object.
We know that for a 3 × 10^(-5) raindrop, the terminal velocity is about 9 m/s. At terminal velocity, the drag force balances the weight of the raindrop, so we can write:
F_d = mg
where m is the mass of the raindrop and g is the acceleration due to gravity (9.8 m/s^2).
Using these equations, we can solve for the value of b:
mg = bv_t^2
b = mg/v_t^2
Plugging in the values given, we get:
b = (3 × 10^(-5) kg)(9.8 m/s^2)/(9 m/s)^2
b ≈ 3.14 × 10^(-5) kg/m
To find the time required for the raindrop to reach 63% of terminal velocity, we can use the following equation:
v(t) = v_t(1 - e^(-kt/m))
where v(t) is the velocity of the raindrop at time t, k is a constant related to b and the density of the fluid (air), and e is the base of the natural logarithm.
At 63% of terminal velocity, v(t) = 0.63v_t. Plugging this into the equation above and solving for t, we get:
t = -m/k * ln(1 - 0.63)
Plugging in the values of m, k, and b, we get:
t = -(3 × 10^(-5) kg)/(0.5 * 1.2 kg/m^3 * 3.14 × 10^(-5) kg/m) * ln(0.37)
t ≈ 0.068 s
Therefore, it would take approximately 0.068 seconds for the raindrop to reach 63% of its terminal velocity.