A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of `2 ms^-2.` The ratio of time of ascent to time of descent 13 (`g = 10 ms^-2`)

Options

A. `1 : 1`
B. `sqrt(2/3)`
C. `2/3`
D. `sqrt(3/2)`

Explanation

To solve the problem, we need to analyze the motion of the particle thrown upwards under the influence of gravity and a constant resistance force.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
- When the particle is thrown upwards, it experiences two forces:
- Gravitational force (downward) = \( mg \)
- Resistance force (downward) = \( R \)
- The total downward force when the particle is moving upwards is \( F_{down} = mg + R \).

2. **Calculate the Retardation:**
- Given that the retardation due to the resistance force is \( 2 \, \text{m/s}^2 \) and \( g = 10 \, \text{m/s}^2 \), we can express the total retardation when moving upwards as:
\[ a_{up} = g + 2 = 10 + 2 = 12 \, \text{m/s}^2 \]

3. **Time of Ascent:**
- Let \( u \) be the initial velocity with which the particle is thrown upwards, and \( t_a \) be the time of ascent. The final velocity at the highest point is \( 0 \).
- Using the equation of motion:
\[ v = u - a_{up} \cdot t_a \]
Setting \( v = 0 \):
\[ 0 = u - 12 t_a \implies t_a = \frac{u}{12} \]

4. **Time of Descent:**
- During descent, the forces acting on the particle are:
- Gravitational force (downward) = \( mg \)
- Resistance force (upward) = \( R \)
- The total downward force when the particle is moving downwards is \( F_{down} = mg - R \).
- The effective acceleration during descent is:
\[ a_{down} = g - 2 = 10 - 2 = 8 \, \text{m/s}^2 \]
- Let \( t_d \) be the time of descent. The particle starts from rest at the highest point and falls under the influence of \( 8 \, \text{m/s}^2 \):
\[ h = \frac{1}{2} a_{down} t_d^2 \implies h = \frac{1}{2} \cdot 8 \cdot t_d^2 = 4 t_d^2 \]

5. **Relate the Heights:**
- The height \( h \) reached during ascent can also be expressed in terms of \( t_a \):
\[ h = u t_a - \frac{1}{2} a_{up} t_a^2 = u \left(\frac{u}{12}\right) - \frac{1}{2} \cdot 12 \left(\frac{u}{12}\right)^2 \]
Simplifying:
\[ h = \frac{u^2}{12} - \frac{1}{2} \cdot 12 \cdot \frac{u^2}{144} = \frac{u^2}{12} - \frac{u^2}{24} = \frac{u^2}{24} \]

6. **Equate Heights:**
- From ascent and descent:
\[ 4 t_d^2 = \frac{u^2}{24} \]
- Now, express \( t_d \) in terms of \( u \):
\[ t_d^2 = \frac{u^2}{96} \implies t_d = \frac{u}{4\sqrt{6}} \]

7. **Find the Ratio of Time of Ascent to Time of Descent:**
- Now we have:
\[ t_a = \frac{u}{12}, \quad t_d = \frac{u}{4\sqrt{6}} \]
- The ratio \( \frac{t_a}{t_d} \) is:
\[ \frac{t_a}{t_d} = \frac{\frac{u}{12}}{\frac{u}{4\sqrt{6}}} = \frac{4\sqrt{6}}{12} = \frac{\sqrt{6}}{3} \]

### Final Answer:
The ratio of time of ascent to time of descent is \( \frac{\sqrt{6}}{3} \).