A particle is projected with velocity `20 ms ^(-1)` at angle `60^@` with horizontal . The radius of curvature of trajectory , at the instant when velocity of projectile become perpendicular to velocity of projection is , `(g=10 ms ^(-1))`

Options

A. `60sqrt(3)m`
B. `80/(sqrt(3))m`
C. `40sqrt(3)m`
D. `80/(3sqrt(3))m`

Answer

To solve the problem, we need to find the radius of curvature of the trajectory of a projectile at the instant when its velocity becomes perpendicular to the initial velocity of projection.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial velocity \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Components of Initial Velocity:**
- The horizontal component of velocity:
\[ u_x = u \cos \theta = 20 \cos 60^\circ = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]
- The vertical component of velocity:
\[ u_y = u \sin \theta = 20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

3. **Determine the Time When Velocity Becomes Perpendicular:**
- At the time \( t \), the vertical component of velocity \( v_y \) can be expressed as:
\[ v_y = u_y - g t = 10\sqrt{3} - 10t \]
- The horizontal component remains constant:
\[ v_x = u_x = 10 \, \text{m/s} \]
- The condition for the velocities to be perpendicular is given by:
\[ u_x v_x + u_y v_y = 0 \]
- Substituting the values:
\[ 10 \cdot 10 + 10\sqrt{3} \cdot (10\sqrt{3} - 10t) = 0 \]
- Simplifying:
\[ 100 + 300 - 100\sqrt{3}t = 0 \implies 400 = 100\sqrt{3}t \implies t = \frac{4}{\sqrt{3}} \, \text{s} \]

4. **Calculate the Velocity at this Time:**
- Substitute \( t \) back into the equations for \( v_x \) and \( v_y \):
\[ v_y = 10\sqrt{3} - 10 \left(\frac{4}{\sqrt{3}}\right) = 10\sqrt{3} - \frac{40}{\sqrt{3}} = \frac{30\sqrt{3}}{3} - \frac{40}{\sqrt{3}} = \frac{30 - 40}{\sqrt{3}} = \frac{-10}{\sqrt{3}} \, \text{m/s} \]
- The magnitude of the velocity \( v \) is:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + \left(-\frac{10}{\sqrt{3}}\right)^2} = \sqrt{100 + \frac{100}{3}} = \sqrt{\frac{300 + 100}{3}} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}} \, \text{m/s} \]

5. **Calculate the Radial Acceleration \( a_r \):**
- The radial acceleration \( a_r \) is given by:
\[ a_r = g \cos \theta \]
- At \( \theta = 30^\circ \) (the angle of the velocity vector):
\[ a_r = g \cos 30^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}^2 \]

6. **Calculate the Radius of Curvature \( R \):**
- The radius of curvature \( R \) is given by:
\[ R = \frac{v^2}{a_r} \]
- Substituting the values:
\[ R = \frac{\left(\frac{20}{\sqrt{3}}\right)^2}{5\sqrt{3}} = \frac{\frac{400}{3}}{5\sqrt{3}} = \frac{400}{15\sqrt{3}} = \frac{80}{3\sqrt{3}} \, \text{m} \]

### Final Answer:
The radius of curvature of the trajectory at the instant when the velocity of the projectile becomes perpendicular to the velocity of projection is:
\[ R = \frac{80}{3\sqrt{3}} \, \text{m} \]