A particle is projected horizontally with a speed `v_(0) ` at t = 0 from a certain point above the ground . What is the magnitude of tangential acceleration of the particle at `t = t_(1)` is ( Assume that the particle does not strike the ground between the time interval t = 0 to `t = t_(1)` )

Options

A. `( g t)/( sqrt(v_(0)^(2)+ g^(2) + 2))`
B. `(tg^(2))/(sqrt(v_(0)^(2) +g^(2) t^(2)))`
C. `(t)/( gsqrt(v_(0)^(2) + g^(2) t^(2)))`
D. `(g t)/( sqrt( g + v_(0)^(2) t))`

Answer

To solve the problem, we need to find the magnitude of the tangential acceleration of a particle projected horizontally with an initial speed \( v_0 \) at time \( t = 0 \). We will analyze the motion of the particle in both horizontal and vertical directions.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is projected horizontally, which means its initial vertical velocity \( u_y = 0 \).
- The horizontal velocity \( u_x = v_0 \) remains constant since there is no horizontal acceleration.

2. **Vertical Motion**:
- The only force acting on the particle in the vertical direction is gravity. Thus, the vertical acceleration \( a_y = g \) (where \( g \) is the acceleration due to gravity).
- The vertical velocity \( v_y \) at time \( t \) can be calculated using the equation of motion:
\[ v_y = u_y + a_y t = 0 + g t = g t \]

3. **Horizontal Motion**:
- The horizontal velocity remains constant:
\[ v_x = v_0 \]

4. **Resultant Velocity**:
- At time \( t \), the resultant velocity \( v \) of the particle can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0)^2 + (g t)^2} \]

5. **Tangential Acceleration**:
- The tangential acceleration is defined as the rate of change of the magnitude of the velocity. Since the horizontal component \( v_x \) is constant and only the vertical component \( v_y \) is changing, we can find the tangential acceleration \( a_t \) as follows:
- The magnitude of the tangential acceleration is given by:
\[ a_t = \frac{d|v|}{dt} \]
- To find \( \frac{d|v|}{dt} \), we differentiate \( v \):
\[ v = \sqrt{(v_0)^2 + (g t)^2} \]
- Using the chain rule:
\[ \frac{d|v|}{dt} = \frac{1}{2\sqrt{(v_0)^2 + (g t)^2}} \cdot (2g t) = \frac{g t}{\sqrt{(v_0)^2 + (g t)^2}} \]

6. **Final Expression**:
- Therefore, the magnitude of the tangential acceleration at time \( t = t_1 \) is:
\[ a_t = \frac{g t_1}{\sqrt{(v_0)^2 + (g t_1)^2}} \]

### Summary:
The magnitude of the tangential acceleration of the particle at time \( t = t_1 \) is given by:
\[ a_t = \frac{g t_1}{\sqrt{(v_0)^2 + (g t_1)^2}} \]