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 find 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 can follow these steps:

### Step 1: Break down the initial velocity into components
The initial velocity \( u = 20 \, \text{m/s} \) can be broken down into horizontal and vertical components using trigonometric functions:
- Horizontal component \( u_x = u \cos \theta = 20 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \)
- Vertical component \( u_y = u \sin \theta = 20 \sin 30^\circ = 20 \times \frac{1}{2} = 10 \, \text{m/s} \)

### Step 2: Determine the position after 1 second
Since there is no acceleration in the horizontal direction, the horizontal distance traveled after 1 second is:
\[ x = u_x \cdot t = 10\sqrt{3} \cdot 1 = 10\sqrt{3} \, \text{m} \]

For the vertical position, we can use the kinematic equation:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ y = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 10 - 5 = 5 \, \text{m} \]

### Step 3: Calculate the velocity components after 1 second
The horizontal component of velocity remains constant:
\[ v_x = u_x = 10\sqrt{3} \, \text{m/s} \]

The vertical component of velocity after 1 second is given by:
\[ v_y = u_y - g t = 10 - 10 \cdot 1 = 0 \, \text{m/s} \]

### Step 4: Calculate the total velocity
The total velocity \( v \) after 1 second can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10\sqrt{3})^2 + 0^2} = 10\sqrt{3} \, \text{m/s} \]

### Step 5: Calculate the radius of curvature
The radius of curvature \( r \) can be found using the formula:
\[ r = \frac{(1 + (dy/dx)^2)^{3/2}}{d^2y/dx^2} \]
We need to find \( dy/dx \) and \( d^2y/dx^2 \).

The trajectory equation of projectile motion is:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]
Substituting \( \theta = 30^\circ \):
\[ y = x \cdot \frac{1}{\sqrt{3}} - \frac{10 x^2}{2 \cdot 400 \cdot \frac{3}{4}} \]
After simplification, we find:
\[ dy/dx = \frac{1}{\sqrt{3}} - \frac{x}{60} \]
At \( x = 10\sqrt{3} \):
\[ dy/dx = \frac{1}{\sqrt{3}} - \frac{10\sqrt{3}}{60} = \frac{1}{\sqrt{3}} - \frac{\sqrt{3}}{6} = \frac{2}{6\sqrt{3}} = \frac{1}{3\sqrt{3}} \]

Now, \( d^2y/dx^2 = -\frac{1}{30} \).

Now substituting these values into the radius of curvature formula:
\[ r = \frac{(1 + (\frac{1}{3\sqrt{3}})^2)^{3/2}}{-\frac{1}{30}} \]
Calculating this gives us the radius of curvature.

### Step 6: Calculate centripetal acceleration
Centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting \( v = 10\sqrt{3} \) and the calculated \( r \):
\[ a_c = \frac{(10\sqrt{3})^2}{r} \]

### Final Result
After calculating, we find that the centripetal acceleration \( a_c \) is \( -10 \, \text{m/s}^2 \) (the negative sign indicates direction).