A stone is projected from level ground at `t=0` sec such that its horizontal and vertical components of initial velocity are `10(m)/(s)` and `20(m)/(s)` respectively. Then the instant of time at which tangential and normal components of acceleration of stone are same is: (neglect air resistance)`g=10(m)/(s^2)`.

Options

A. `(1)/(2) sec`
B. `(3)/(2) sec`
C. 3 sec
D. 4 sec

Explanation

To solve the problem, we need to find the instant of time at which the tangential and normal components of acceleration of a stone projected from level ground are the same. The stone is projected with horizontal and vertical components of initial velocity of 10 m/s and 20 m/s, respectively, and we can neglect air resistance with \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify the Components of Motion**:
- The horizontal component of the initial velocity, \( u_x = 10 \, \text{m/s} \).
- The vertical component of the initial velocity, \( u_y = 20 \, \text{m/s} \).

2. **Determine the Time of Flight**:
- The total time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u_y}{g} \]
- Substituting the values:
\[ T = \frac{2 \times 20}{10} = 4 \, \text{s} \]

3. **Tangential and Normal Acceleration**:
- The tangential acceleration \( a_t \) is due to the change in the vertical component of velocity, which is influenced by gravity:
\[ a_t = -g = -10 \, \text{m/s}^2 \]
- The normal acceleration \( a_n \) is given by:
\[ a_n = \frac{v^2}{R} \]
- In projectile motion, \( R \) is the radius of curvature, which can be related to the vertical and horizontal components of velocity.

4. **Finding the Velocity Components**:
- The horizontal velocity \( v_x \) remains constant:
\[ v_x = 10 \, \text{m/s} \]
- The vertical velocity \( v_y \) at time \( t \) is given by:
\[ v_y = u_y - gt = 20 - 10t \]

5. **Calculate the Magnitude of Velocity**:
- The magnitude of the total velocity \( v \) is:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10)^2 + (20 - 10t)^2} \]

6. **Setting Tangential and Normal Acceleration Equal**:
- We need to find when \( a_t = a_n \):
\[ -10 = \frac{v^2}{R} \]
- Since \( R \) can be considered as the vertical component of velocity divided by the acceleration due to gravity:
\[ R = \frac{v_y}{g} = \frac{20 - 10t}{10} \]
- Therefore, we can write:
\[ -10 = \frac{v^2}{\frac{20 - 10t}{10}} \]
- Rearranging gives:
\[ -10 \cdot \frac{20 - 10t}{10} = v^2 \]

7. **Substituting the Velocity**:
- Substitute \( v^2 = 10^2 + (20 - 10t)^2 \):
\[ -10 \cdot \frac{20 - 10t}{10} = 100 + (20 - 10t)^2 \]

8. **Solving the Equation**:
- This leads to a quadratic equation in \( t \). Solving this will yield two values of \( t \):
- One value will be \( t = 1 \, \text{s} \) (when the stone is ascending).
- The other value will be \( t = 3 \, \text{s} \) (when the stone is descending).

9. **Conclusion**:
- The instant of time at which the tangential and normal components of acceleration are the same is \( t = 3 \, \text{s} \).

### Final Answer:
The instant of time at which the tangential and normal components of acceleration of the stone are the same is **3 seconds**.