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 of finding the magnitude of tangential acceleration of a particle projected horizontally with speed \( v_0 \) at \( t = 0 \), we can follow these steps:

### Step 1: Understand the Motion
The particle is projected horizontally, meaning its initial vertical velocity \( V_{y0} = 0 \). The only force acting on the particle in the vertical direction is gravity, which causes a downward acceleration of \( g \).

### Step 2: Determine the Vertical Velocity at Time \( t_1 \)
The vertical velocity \( V_y \) at time \( t_1 \) can be calculated using the equation of motion:
\[ V_y = V_{y0} + a_y \cdot t \]
Since \( V_{y0} = 0 \) and \( a_y = g \) (downward), we have:
\[ V_y = 0 + g \cdot t_1 = g t_1 \]

### Step 3: Determine the Angle of the Velocity Vector
At time \( t_1 \), the horizontal velocity \( V_x \) remains constant and equal to \( v_0 \). The vertical velocity is \( V_y = g t_1 \). The angle \( \theta \) of the resultant velocity vector can be found using the tangent function:
\[ \tan \theta = \frac{V_y}{V_x} = \frac{g t_1}{v_0} \]

### Step 4: Calculate \( \sin \theta \)
To find \( \sin \theta \), we can use the relationship:
\[ \sin \theta = \frac{V_y}{\sqrt{V_x^2 + V_y^2}} = \frac{g t_1}{\sqrt{v_0^2 + (g t_1)^2}} \]

### Step 5: Determine the Tangential Acceleration
The tangential acceleration \( a_t \) is given by:
\[ a_t = g \sin \theta \]
Substituting for \( \sin \theta \):
\[ a_t = g \cdot \frac{g t_1}{\sqrt{v_0^2 + (g t_1)^2}} = \frac{g^2 t_1}{\sqrt{v_0^2 + (g t_1)^2}} \]

### Final Result
Thus, the magnitude of the tangential acceleration of the particle at time \( t_1 \) is:
\[ a_t = \frac{g^2 t_1}{\sqrt{v_0^2 + (g t_1)^2}} \]