An object is projected from ground with speed 20 m/s at angle `30^(@)` with horizontal. Its centripetal acceleration one second after the projection is
[ Take g = 10 m/`s^(2)` ]

Options

A. 10 `m//s^(2)`
B. zero
C. 5 `m//s^(2)`
D. 12 `m//s^(2)`

Answer

To solve the problem of finding the centripetal acceleration of an object projected from the ground with a speed of 20 m/s at an angle of 30 degrees with the horizontal, we will follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( u \) can be broken down into horizontal and vertical components using trigonometric functions:
- Horizontal component, \( u_x = u \cos(\theta) \)
- Vertical component, \( u_y = u \sin(\theta) \)

Given:
- \( u = 20 \, \text{m/s} \)
- \( \theta = 30^\circ \)

Calculating the components:
\[ u_x = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]
\[ u_y = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

### Step 2: Determine the velocity at \( t = 1 \, \text{s} \)
At \( t = 1 \, \text{s} \), the vertical velocity can be calculated using the equation of motion:
\[ v_y = u_y - g t \]
Where \( g = 10 \, \text{m/s}^2 \):
\[ v_y = 10 - 10 \times 1 = 0 \, \text{m/s} \]

The horizontal velocity remains constant:
\[ v_x = u_x = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Determine the total velocity at \( t = 1 \, \text{s} \)
The total velocity \( v \) at \( t = 1 \, \text{s} \) is:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10\sqrt{3})^2 + 0^2} = 10\sqrt{3} \, \text{m/s} \]

### Step 4: Determine the centripetal acceleration
Centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
However, since we are looking for the centripetal acceleration at the highest point, we note that at this point, the only acceleration acting on the object is due to gravity, which acts downwards. Thus, the centripetal acceleration is equal to the gravitational acceleration:
\[ a_c = g = 10 \, \text{m/s}^2 \]

### Final Answer
The centripetal acceleration one second after the projection is:
\[ \boxed{10 \, \text{m/s}^2} \]