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 find the time taken by the ball to rise to its zenith when thrown upward with an initial velocity \( V_0 \) and affected by a drag force \( mg\gamma v^2 \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Ball
When the ball is thrown upwards, it experiences two forces:
1. The gravitational force acting downwards: \( F_g = mg \)
2. The drag force acting downwards due to its velocity: \( F_d = mg\gamma v^2 \)

### Step 2: Write the Equation of Motion
The net force acting on the ball can be expressed as:
\[ F_{net} = -mg - mg\gamma v^2 = -m(g + \gamma v^2) \]
Using Newton's second law \( F = ma \), we can write:
\[ ma = -m(g + \gamma v^2) \]
Dividing through by \( m \) gives us the acceleration \( a \):
\[ a = -g - \gamma v^2 \]

### Step 3: Relate Acceleration to Velocity
Acceleration can also be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = -g - \gamma v^2 \]

### Step 4: Rearrange the Equation
Rearranging gives:
\[ \frac{dv}{g + \gamma v^2} = -dt \]

### Step 5: Integrate Both Sides
Now we will integrate both sides. The left side will be integrated with respect to \( v \) from \( V_0 \) to \( 0 \) (the zenith) and the right side with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_{V_0}^{0} \frac{dv}{g + \gamma v^2} = -\int_{0}^{t} dt \]

### Step 6: Solve the Integral
The integral on the left can be solved using the formula:
\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \]
In our case, \( a^2 = \frac{g}{\gamma} \) so \( a = \sqrt{\frac{g}{\gamma}} \):
\[ \int_{V_0}^{0} \frac{dv}{g + \gamma v^2} = \frac{1}{\sqrt{g\gamma}} \left[ \tan^{-1}\left(\frac{v \sqrt{\gamma}}{\sqrt{g}}\right) \right]_{V_0}^{0} \]
Evaluating this gives:
\[ \frac{1}{\sqrt{g\gamma}} \left( \tan^{-1}(0) - \tan^{-1}\left(\frac{V_0 \sqrt{\gamma}}{\sqrt{g}}\right) \right) = -\frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{V_0 \sqrt{\gamma}}{\sqrt{g}}\right) \]

### Step 7: Equate and Solve for Time \( t \)
Setting the integrals equal gives:
\[ -\frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{V_0 \sqrt{\gamma}}{\sqrt{g}}\right) = -t \]
Thus, we can express the time taken to rise to the zenith as:
\[ t = \frac{1}{\sqrt{g\gamma}} \tan^{-1}\left(\frac{V_0 \sqrt{\gamma}}{\sqrt{g}}\right) \]

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