A particle is projected up a smooth inclined plane of inclination `60^(@)` along the line of greatest slope. If it comes to instantaneous rest after 2 seconds, then the velocity of projection is (`g = 9.8 m//"sec"^(2)`)

Options

A. `9.8 m//s`
B. `10 m//s`
C. `16.97 m//s`
D. `19.6 m//s`

Answer

To solve the problem, we need to find the initial velocity of a particle projected up a smooth inclined plane at an angle of \(60^\circ\) that comes to rest after 2 seconds. We will use the equations of motion and the components of gravitational acceleration along the incline.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Inclination angle, \(\theta = 60^\circ\)
- Time of ascent, \(t = 2 \, \text{s}\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

2. **Determine the Acceleration along the Incline:**
- The acceleration acting on the particle along the incline is due to the component of gravitational force acting down the incline. This can be calculated as:
\[ a = g \sin(\theta) = 9.8 \sin(60^\circ) \]
- Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ a = 9.8 \times \frac{\sqrt{3}}{2} \approx 8.49 \, \text{m/s}^2 \]

3. **Use the Equation of Motion:**
- We will use the first equation of motion:
\[ v = u + at \]
- Here, \(v\) is the final velocity (which is 0 when the particle comes to rest), \(u\) is the initial velocity, \(a\) is the acceleration (which will be negative since it is acting opposite to the direction of motion), and \(t\) is the time.
- Substituting the known values:
\[ 0 = u - 8.49 \times 2 \]

4. **Solve for Initial Velocity \(u\):**
- Rearranging the equation gives:
\[ u = 8.49 \times 2 \]
- Calculating this:
\[ u = 16.98 \, \text{m/s} \]

5. **Final Result:**
- The initial velocity of projection is approximately \(16.98 \, \text{m/s}\).

### Summary:
The velocity of projection of the particle is approximately \(16.98 \, \text{m/s}\).