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 velocity of projection. Let's break down the solution step by step.

### Step 1: Determine the initial velocity components
The initial velocity \( u \) is given as \( 20 \, \text{m/s} \) at an angle of \( 60^\circ \) with the horizontal.

- The horizontal component of the initial velocity \( u_x \):
\[ u_x = u \cos(60^\circ) = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

- The vertical component of the initial velocity \( u_y \):
\[ u_y = u \sin(60^\circ) = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 2: Determine the condition for perpendicular velocities
The velocity of the projectile becomes perpendicular to the initial velocity when the angle of the velocity vector is \( 30^\circ \) with the horizontal (downward).

### Step 3: Find the velocity at that instant
At this position, the horizontal component of the velocity remains unchanged:
\[ v_x = u_x = 10 \, \text{m/s} \]

Let \( v \) be the magnitude of the velocity at this point. The vertical component of the velocity \( v_y \) can be determined using the angle:
\[ \tan(30^\circ) = \frac{v_y}{v_x} \implies v_y = v_x \tan(30^\circ) = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s} \]

Now, we can find the magnitude of the velocity \( v \):
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + \left(\frac{10}{\sqrt{3}}\right)^2} = \sqrt{100 + \frac{100}{3}} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}} \, \text{m/s} \]

### Step 4: Calculate the acceleration components
The acceleration in the vertical direction is due to gravity \( g = 10 \, \text{m/s}^2 \). The component of acceleration perpendicular to the velocity vector is:
\[ a_{\perp} = g \cos(30^\circ) = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}^2 \]

### Step 5: Calculate the radius of curvature
The radius of curvature \( R \) is given by the formula:
\[ R = \frac{v^2}{a_{\perp}} \]
Substituting the values we found:
\[ R = \frac{\left(\frac{20}{\sqrt{3}}\right)^2}{5\sqrt{3}} = \frac{\frac{400}{3}}{5\sqrt{3}} = \frac{400}{15\sqrt{3}} = \frac{80}{3\sqrt{3}} \, \text{m} \]

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