A particle is projected from ground with an initial velocity `20m//sec` making an angle `60^(@)` with horizontal . If `R_(1)` and `R_(2)` are radius of curvatures of the particle at point of projection and highest point respectively, then find the value of `(R_(1))/(R_(2))`.

Explanation

To solve the problem, we need to find the ratio of the radii of curvature \( R_1 \) and \( R_2 \) at the point of projection and at the highest point of the projectile motion, respectively.

### Step 1: Identify the given data
- Initial velocity \( V_0 = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (we can use \( g \) as \( 9.81 \, \text{m/s}^2 \) for more precision, but it won't affect the ratio)

### Step 2: Calculate the radius of curvature at the point of projection \( R_1 \)
At the point of projection, the velocity \( V \) is \( V_0 \). The components of gravitational acceleration are:
- \( g \cos \theta \) (perpendicular to the velocity)
- \( g \sin \theta \) (along the direction of velocity)

The radius of curvature \( R_1 \) is given by the formula:
\[ R_1 = \frac{V_0^2}{g \cos \theta} \]

Substituting the values:
\[ R_1 = \frac{(20)^2}{g \cos(60^\circ)} \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ R_1 = \frac{400}{g \cdot \frac{1}{2}} = \frac{800}{g} \]

### Step 3: Calculate the radius of curvature at the highest point \( R_2 \)
At the highest point, the horizontal component of the velocity is:
\[ V = V_0 \cos \theta \]
Thus,
\[ V = 20 \cdot \cos(60^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

The radius of curvature \( R_2 \) at the highest point is given by:
\[ R_2 = \frac{(V_0 \cos \theta)^2}{g} \]
Substituting the values:
\[ R_2 = \frac{(10)^2}{g} = \frac{100}{g} \]

### Step 4: Find the ratio \( \frac{R_1}{R_2} \)
Now we can find the ratio of the two radii of curvature:
\[ \frac{R_1}{R_2} = \frac{\frac{800}{g}}{\frac{100}{g}} = \frac{800}{100} = 8 \]

### Final Answer
Thus, the value of \( \frac{R_1}{R_2} = 8 \).

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