A particle is projected vertically upwards with speed `v_(0)`. The drag force acting on it given by `f_("drag")=mlambdav^(2)`. The time when it is at maximum height is :

Options

A. `(1)/(sqrt(glamda))tan^(-1)((sqrt(lamda))/(sqrt(g))v_(0))`
B. `sqrt(glamda)tan^(-1)((sqrt(lamda))/(sqrt(g))v_(0))`
C. `(tan^(-1))/(sqrt(glamda))((sqrt(g))/(sqrt(lamda))v_(0))`
D. `(1)/(sqrt(glamda))tan^(-1)((sqrt(lamda))/(sqrt(g))(1)/(v_(0)))`

Answer

To solve the problem of finding the time when a particle projected vertically upwards with an initial speed \( v_0 \) reaches its maximum height while experiencing a drag force given by \( f_{\text{drag}} = m\lambda v^2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
When the particle is projected upwards, two forces act on it:
- The gravitational force \( F_g = mg \) acting downwards.
- The drag force \( F_{\text{drag}} = m\lambda v^2 \) also acting downwards.

2. **Write the Net Force Equation:**
The net force \( F_{\text{net}} \) acting on the particle can be expressed as:
\[ F_{\text{net}} = -mg - m\lambda v^2 \]
According to Newton's second law, this net force is equal to the mass times the acceleration \( a \):
\[ ma = -mg - m\lambda v^2 \]
Dividing through by \( m \) gives:
\[ a = -g - \lambda v^2 \]

3. **Relate Acceleration to Velocity:**
We know that acceleration \( a \) can also be expressed as the time derivative of velocity:
\[ a = \frac{dv}{dt} \]
Therefore, we can rewrite our equation as:
\[ \frac{dv}{dt} = -g - \lambda v^2 \]

4. **Separate Variables:**
Rearranging the equation allows us to separate variables:
\[ \frac{dv}{g + \lambda v^2} = -dt \]

5. **Integrate Both Sides:**
We need to integrate both sides. The limits for \( v \) will be from \( v_0 \) to \( 0 \) (as the particle comes to rest at maximum height), and for \( t \) from \( 0 \) to \( t \):
\[ \int_{v_0}^{0} \frac{dv}{g + \lambda v^2} = -\int_{0}^{t} dt \]

6. **Evaluate the Left Integral:**
The left integral can be solved using a trigonometric substitution or recognizing it as a standard integral. The result is:
\[ \frac{1}{\sqrt{\lambda g}} \tan^{-1}\left(\frac{v}{\sqrt{\frac{g}{\lambda}}}\right) \bigg|_{v_0}^{0} = -t \]
Evaluating this gives:
\[ \frac{1}{\sqrt{\lambda g}} \left( \tan^{-1}(0) - \tan^{-1}\left(\frac{v_0}{\sqrt{\frac{g}{\lambda}}}\right) \right) = -t \]
Since \( \tan^{-1}(0) = 0 \), we have:
\[ -\frac{1}{\sqrt{\lambda g}} \tan^{-1}\left(\frac{v_0}{\sqrt{\frac{g}{\lambda}}}\right) = -t \]
Thus, we can express \( t \) as:
\[ t = \frac{1}{\sqrt{\lambda g}} \tan^{-1}\left(\frac{v_0}{\sqrt{\frac{g}{\lambda}}}\right) \]

### Final Answer:
The time when the particle is at maximum height is given by:
\[ t = \frac{1}{\sqrt{\lambda g}} \tan^{-1}\left(\frac{v_0}{\sqrt{\frac{g}{\lambda}}}\right) \]