A heavy particle is projected from a point on the horizontal at an angle `60^(@)` with the horizontal with a speed of `10m//s`. Then the radius of the curvature of its path at the instant of crossing the same horizontal is `……………………..`.

Explanation

To find the radius of curvature of the path of a heavy particle projected at an angle of \(60^\circ\) with an initial speed of \(10 \, \text{m/s}\), we will follow these steps:

### Step 1: Analyze the motion of the particle
The particle is projected at an angle of \(60^\circ\) with the horizontal. The initial speed is \(10 \, \text{m/s}\). As the particle moves, it will follow a parabolic trajectory due to the influence of gravity.

### Step 2: Determine the components of the initial velocity
The initial velocity can be broken down into horizontal and vertical components:
- \(V_x = V \cdot \cos(\theta) = 10 \cdot \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s}\)
- \(V_y = V \cdot \sin(\theta) = 10 \cdot \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}\)

### Step 3: Analyze the motion at the instant of crossing the horizontal
At the instant the particle crosses the horizontal plane again, its vertical velocity component will be equal in magnitude to its initial vertical velocity but directed downwards. Thus, the vertical component of the velocity at this point is:
- \(V_y = 5\sqrt{3} \, \text{m/s}\)

The horizontal component remains unchanged:
- \(V_x = 5 \, \text{m/s}\)

### Step 4: Calculate the total velocity at the horizontal crossing
The total velocity \(V\) at the instant of crossing the horizontal can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(5)^2 + (5\sqrt{3})^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 5: Determine the normal acceleration
The normal acceleration \(a_n\) at this point can be expressed as:
\[ a_n = g \cos(\theta) \]
where \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)) and \(\theta\) is the angle of projection. Thus:
\[ a_n = 10 \cdot \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s}^2 \]

### Step 6: Calculate the radius of curvature
The radius of curvature \(r\) can be calculated using the formula:
\[ r = \frac{V^2}{a_n} \]
Substituting the values:
\[ r = \frac{(10)^2}{5} = \frac{100}{5} = 20 \, \text{m} \]

### Final Answer
The radius of curvature of the particle's path at the instant of crossing the same horizontal is \(20 \, \text{m}\).

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