Two spherical raindrops of equal size are falling vertically through air with a velocity of `1 m//s`. What would be the terminal speed if these two drops were to coalesce to form a large spherical drop?

Explanation

To solve the problem of finding the terminal speed of a larger spherical raindrop formed by the coalescence of two smaller raindrops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two spherical raindrops of equal size falling with a velocity of 1 m/s. When they coalesce, they form a larger spherical raindrop. We need to find the terminal speed of this larger drop.

2. **Define Variables**:
- Let the radius of each smaller raindrop be \( r \).
- The radius of the larger raindrop formed by coalescing the two smaller drops will be \( R \).
- The terminal speed of the smaller drops is \( v_t = 1 \, \text{m/s} \).
- We need to find the terminal speed of the larger drop, denoted as \( v'_t \).

3. **Volume Conservation**:
The volume of the two smaller raindrops combined is equal to the volume of the larger raindrop:
\[ 2 \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3 \]
Simplifying, we find:
\[ 2r^3 = R^3 \]
Therefore, we can express \( R \) in terms of \( r \):
\[ R = (2)^{1/3} r \]

4. **Terminal Velocity Relation**:
The terminal velocity \( v_t \) of a spherical object falling through a fluid is proportional to the square of its radius:
\[ v_t \propto r^2 \]
Thus, we can write:
\[ \frac{v'_t}{v_t} = \left( \frac{R}{r} \right)^2 \]

5. **Substituting for \( R \)**:
Substitute \( R = (2)^{1/3} r \) into the equation:
\[ \frac{v'_t}{v_t} = \left( \frac{(2)^{1/3} r}{r} \right)^2 = (2)^{2/3} \]

6. **Calculating \( v'_t \)**:
Now, we can express \( v'_t \):
\[ v'_t = v_t \cdot (2)^{2/3} \]
Substituting \( v_t = 1 \, \text{m/s} \):
\[ v'_t = 1 \cdot (2)^{2/3} \]

7. **Finding the Numerical Value**:
Calculate \( (2)^{2/3} \):
\[ (2)^{2/3} \approx 1.5874 \]
Therefore, the terminal speed of the larger drop is approximately:
\[ v'_t \approx 1.587 \, \text{m/s} \]

### Final Answer:
The terminal speed of the larger raindrop formed by the coalescence of the two smaller raindrops is approximately **1.587 m/s**.