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 the ground are equal.

### Step-by-Step Solution:

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

2. **Understand the Motion:**
- The stone is projected at an angle, and its motion can be analyzed in two dimensions: horizontal and vertical.
- The horizontal motion is uniform since there is no horizontal acceleration (neglecting air resistance).
- The vertical motion is subject to gravitational acceleration, \( g = 10 \, \text{m/s}^2 \).

3. **Tangential and Normal Components of Acceleration:**
- The tangential component of acceleration (\( a_t \)) is due to the change in the vertical velocity and is equal to \( -g \) (since gravity acts downward).
- The normal component of acceleration (\( a_n \)) is associated with the change in direction of the velocity vector.

4. **Finding the Angle \( \theta \):**
- At any point in time, the angle \( \theta \) of the velocity vector can be determined from the horizontal and vertical components:
\[ \tan(\theta) = \frac{v_y}{v_x} \]
- The condition for tangential and normal components of acceleration to be equal is:
\[ a_t = a_n \]
- This means:
\[ g \sin(\theta) = g \cos(\theta) \]
- Simplifying gives:
\[ \tan(\theta) = 1 \implies \theta = 45^\circ \]

5. **Finding the Time When \( \theta = 45^\circ \):**
- For \( \theta = 45^\circ \):
\[ v_y = v_x \]
- The horizontal velocity \( v_x \) remains constant at \( 10 \, \text{m/s} \).
- The vertical velocity \( v_y \) can be calculated using the kinematic equation:
\[ v_y = u_y - g t \]
- Setting \( v_y = v_x \):
\[ 10 = 20 - 10 t \]
- Rearranging gives:
\[ 10 t = 10 \implies t = 1 \, \text{s} \]

6. **Total Time of Flight:**
- The total time of flight for the projectile can be calculated as:
\[ T = \frac{2 u_y}{g} = \frac{2 \times 20}{10} = 4 \, \text{s} \]

7. **Finding the Time at the Symmetrical Point:**
- The time at which the tangential and normal components are equal occurs at two points during the projectile's flight: once while ascending and once while descending.
- The time at the descending point is:
\[ T - t = 4 - 1 = 3 \, \text{s} \]

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