A body of mass m is thrown straight up with velocity `v_(0)(v_(0) lt lt v_(e))` Find the velocity `v` with which the body comes down if the air drag equals `kv^(2)`, where `k` is a positive constant and `v` is the velocity of the body.

Explanation

To solve the problem, we need to analyze the motion of a body thrown upwards and then coming down under the influence of gravity and air drag. Here are the steps to derive the expression for the velocity \( v \) with which the body comes down:

### Step 1: Analyze the upward motion
When the body is thrown upwards with an initial velocity \( v_0 \), it experiences two forces:
1. The gravitational force acting downward, \( mg \).
2. The air drag acting downward, which is given by \( kv^2 \).

Using Newton's second law, we can write the equation of motion for the upward journey:
\[ m \frac{dv}{dt} = -mg - kv^2 \]
This can be rearranged to:
\[ \frac{dv}{dt} = -g - \frac{k}{m}v^2 \]

### Step 2: Integrate the upward motion
To find the maximum height \( h \), we can use the relationship between velocity and displacement. We can rewrite the equation as:
\[ m v \frac{dv}{ds} = -mg - kv^2 \]
Rearranging gives:
\[ m v \frac{dv}{ds} = -mg - kv^2 \]
Integrating from \( v_0 \) to \( 0 \) (the maximum height) gives:
\[ \int_{v_0}^{0} v \, dv = -\int_{0}^{h} (g + \frac{k}{m}v^2) \, ds \]

### Step 3: Solve for maximum height
After integrating and simplifying, we find the maximum height \( h \):
\[ h = \frac{m}{2k} \ln\left(1 + \frac{kv_0^2}{mg}\right) \]

### Step 4: Analyze the downward motion
When the body falls back down, the forces acting on it are:
1. The gravitational force \( mg \) acting downward.
2. The air drag \( kv^2 \) acting upward.

The net force equation during the downward motion is:
\[ m \frac{dv}{dt} = mg - kv^2 \]
This can be rearranged to:
\[ \frac{dv}{dt} = g - \frac{k}{m}v^2 \]

### Step 5: Integrate the downward motion
Using the same approach as before, we can write:
\[ m v \frac{dv}{ds} = mg - kv^2 \]
Rearranging gives:
\[ \frac{dv}{g - \frac{k}{m}v^2} = \frac{ds}{m} \]
Integrating from \( 0 \) to \( v \) (the final velocity when it hits the ground) gives:
\[ \int_{0}^{v} \frac{dv}{g - \frac{k}{m}v^2} = \frac{1}{m} \int_{0}^{h} ds \]

### Step 6: Solve for final velocity
After integrating and applying limits, we can derive the expression for the velocity \( v \) when the body comes down:
\[ v = \frac{v_0}{\sqrt{1 + \frac{kv_0^2}{mg}}} \]

### Final Answer
The velocity \( v \) with which the body comes down is:
\[ v = \frac{v_0}{\sqrt{1 + \frac{kv_0^2}{mg}}} \]