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 time at which the tangential and normal components of acceleration of a stone projected at an angle are equal. Let's break down the solution step by step.

### Step 1: Understand the motion components
The stone is projected with:
- Horizontal component of initial velocity, \( u_x = 10 \, \text{m/s} \)
- Vertical component of initial velocity, \( u_y = 20 \, \text{m/s} \)

The acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \), acts downwards.

### Step 2: Identify the components of acceleration
In projectile motion:
- The tangential component of acceleration \( a_t \) is due to the change in the vertical velocity, which is affected by gravity.
- The normal component of acceleration \( a_n \) is due to the change in direction of the velocity vector.

Since there is no horizontal acceleration, the horizontal component of velocity \( V_x \) remains constant at \( 10 \, \text{m/s} \).

### Step 3: Determine the angle of projection
The angle \( \theta \) of the velocity vector can be determined using the initial velocity components:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{20}{10} = 2 \quad \Rightarrow \quad \theta = \tan^{-1}(2) \]

### Step 4: Set up the equations for tangential and normal acceleration
At any time \( t \):
- The vertical component of velocity \( V_y \) can be expressed as:
\[ V_y = u_y - g t = 20 - 10t \]

The tangential acceleration \( a_t \) is given by:
\[ a_t = g \sin \theta \]
And the normal acceleration \( a_n \) is given by:
\[ a_n = g \cos \theta \]

### Step 5: Equate tangential and normal components of acceleration
We need to find the time \( t \) when \( a_t = a_n \):
\[ g \sin \theta = g \cos \theta \]
This simplifies to:
\[ \tan \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 6: Find the time when the angle is 45 degrees
Using the relationship between the vertical and horizontal components of velocity:
\[ \tan \theta = \frac{V_y}{V_x} \]
At \( \theta = 45^\circ \):
\[ V_y = V_x \quad \Rightarrow \quad 20 - 10t = 10 \]
Solving for \( t \):
\[ 20 - 10t = 10 \quad \Rightarrow \quad 10 = 10t \quad \Rightarrow \quad t = 1 \, \text{s} \]

### Step 7: Calculate total time of flight
The total time of flight \( T \) for the projectile can be calculated using:
\[ T = \frac{2u_y}{g} = \frac{2 \times 20}{10} = 4 \, \text{s} \]

### Step 8: Determine the time when tangential and normal accelerations are equal
Since the projectile takes 4 seconds to reach the ground, and we found that \( t = 1 \, \text{s} \) corresponds to the time when \( a_t = a_n \), we need to find the total time from the start to this point:
- The time taken to reach the point where \( a_t = a_n \) is \( 1 \, \text{s} \).
- The time to reach the maximum height (where \( V_y = 0 \)) is \( 2 \, \text{s} \) (half of the total time of flight).

Thus, the time at which the tangential and normal components of acceleration are equal is:
\[ t = 3 \, \text{s} \]

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

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