A stone is thrown horizontally with a velocity of `10m//sec`. Find the radius of curvature of it’s trajectory at the end of `3 s` after motion began. `(g=10m//s^(2))`

Options

A. `10sqrt(10)m`
B. `100sqrt(10)m`
C. `sqrt(10)m`
D. `100m`

Explanation

To find the radius of curvature of the trajectory of a stone thrown horizontally with a velocity of 10 m/s after 3 seconds, we can follow these steps:

### Step 1: Determine the vertical velocity (Vy) after 3 seconds
The vertical velocity can be calculated using the equation of motion:
\[ Vy = u + at \]
Where:
- \( u = 0 \) (initial vertical velocity)
- \( a = g = 10 \, m/s^2 \) (acceleration due to gravity)
- \( t = 3 \, s \)

Substituting the values:
\[ Vy = 0 + (10 \, m/s^2)(3 \, s) = 30 \, m/s \]

### Step 2: Determine the horizontal velocity (Vx)
Since the stone is thrown horizontally, the horizontal velocity remains constant:
\[ Vx = 10 \, m/s \]

### Step 3: Calculate the resultant velocity (Vt)
The resultant velocity \( Vt \) can be calculated using the Pythagorean theorem:
\[ Vt = \sqrt{Vx^2 + Vy^2} \]
Substituting the values:
\[ Vt = \sqrt{(10 \, m/s)^2 + (30 \, m/s)^2} = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10} \, m/s \]

### Step 4: Determine the normal acceleration (An)
The only acceleration acting on the stone is due to gravity, which acts vertically downward. The normal component of acceleration \( An \) can be calculated as:
\[ An = g \cdot \cos(\theta) \]
Where \( \theta \) is the angle of the resultant velocity with respect to the horizontal. We can find \( \cos(\theta) \) using:
\[ \cos(\theta) = \frac{Vx}{Vt} = \frac{10}{10\sqrt{10}} = \frac{1}{\sqrt{10}} \]
Thus,
\[ An = g \cdot \cos(\theta) = 10 \cdot \frac{1}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10} \, m/s^2 \]

### Step 5: Calculate the radius of curvature (R)
The radius of curvature \( R \) can be calculated using the formula:
\[ R = \frac{Vt^2}{An} \]
Substituting the values we have:
\[ R = \frac{(10\sqrt{10})^2}{\sqrt{10}} = \frac{1000}{\sqrt{10}} = 100\sqrt{10} \, m \]

### Final Answer
The radius of curvature of the stone's trajectory at the end of 3 seconds is:
\[ R = 100\sqrt{10} \, m \]