A ball is thrown upward with an initial velocity `V_0` from the surface of the earth.The motion of the ball is affected by a drag force equal to `mgammav^2` (where m is mass of the ball, v is its instantaeous velocity and `gamma` is a constant).Time taken by the ball to rise to its zenith is :

Options

A. `1/sqrt(gammag)tan^(-1)(sqrt(gamma/g)V_0)`
B. `1/sqrt(gammag)sin^(-1)(sqrt(gamma/g)V_0)`
C. `1/sqrt(gammag)"In"(1+sqrt(gamma/g)V_0)`
D. `1/sqrt(2gammag)tan^(-1)(sqrt((2gamma)/g)V_0)`

Answer

To solve the problem of a ball thrown upward with an initial velocity \( V_0 \) while experiencing a drag force proportional to the square of its velocity, we can follow these steps:

### Step 1: Understand the Forces Acting on the Ball
The forces acting on the ball are:
1. The gravitational force downward: \( F_g = mg \)
2. The drag force upward: \( F_d = -m\gamma v^2 \)

The net force \( F \) acting on the ball can be expressed as:
\[ F = -mg - m\gamma v^2 \]

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force is equal to the mass of the ball times its acceleration:
\[ m \frac{dv}{dt} = -mg - m\gamma v^2 \]
Dividing through by \( m \):
\[ \frac{dv}{dt} = -g - \gamma v^2 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{dv}{dt} = -g - \gamma v^2 \]
This can be rewritten as:
\[ \frac{dv}{-g - \gamma v^2} = dt \]

### Step 4: Separate Variables
We will separate the variables \( v \) and \( t \):
\[ \int \frac{dv}{-g - \gamma v^2} = \int dt \]

### Step 5: Integrate Both Sides
The left side requires a specific integration technique. Recognizing that the integral resembles the form of \( \frac{1}{a^2 + x^2} \):
\[ \int \frac{dv}{-g - \gamma v^2} = -\frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{v\sqrt{\gamma}}{\sqrt{g}}\right) + C \]
The right side integrates to:
\[ t + C' \]

### Step 6: Apply Limits
We apply the limits for the initial conditions:
- At \( t = 0 \), \( v = V_0 \)
- At \( t = t_{max} \), \( v = 0 \)

Substituting these limits into the integrated equation gives:
\[ -\frac{1}{\sqrt{g\gamma}} \left(\tan^{-1}(0) - \tan^{-1}\left(\frac{V_0\sqrt{\gamma}}{\sqrt{g}}\right)\right) = t_{max} \]

### Step 7: Solve for \( t_{max} \)
This simplifies to:
\[ t_{max} = \frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{V_0\sqrt{\gamma}}{\sqrt{g}}\right) \]

### Final Answer
Thus, the time taken by the ball to rise to its zenith is:
\[ t_{max} = \frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{V_0\sqrt{\gamma}}{\sqrt{g}}\right) \]

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