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 `……………………..`.

Options

A. Infinite
B. 10 m
C. 11.54 m
D. 20 m

Explanation

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

### Step 1: Determine the components of velocity
The initial velocity \(u\) is given as \(10 \, \text{m/s}\). We can resolve this velocity into horizontal and vertical components:
- Horizontal component:
\[ u_x = u \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]
- Vertical component:
\[ u_y = u \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 2: Determine the velocity at the point of crossing the horizontal
At the instant the particle crosses the horizontal again, the vertical component of the velocity will be equal in magnitude but opposite in direction to the initial vertical component. Therefore, the vertical component of the velocity \(v_y\) at this point is:
\[ v_y = -u_y = -5\sqrt{3} \, \text{m/s} \]
The horizontal component remains the same:
\[ v_x = 5 \, \text{m/s} \]

### Step 3: Calculate the total velocity at the point of 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 4: Determine the centripetal acceleration
The centripetal acceleration \(a_c\) can be expressed in terms of gravitational acceleration \(g\) and the angle of the trajectory. At the point of crossing the horizontal, the effective centripetal acceleration is given by:
\[ a_c = g \cos(60^\circ) = g \cdot \frac{1}{2} \]
Assuming \(g = 10 \, \text{m/s}^2\):
\[ a_c = 10 \cdot \frac{1}{2} = 5 \, \text{m/s}^2 \]

### Step 5: Relate centripetal acceleration to radius of curvature
The centripetal acceleration can also be expressed as:
\[ a_c = \frac{v^2}{R} \]
Where \(R\) is the radius of curvature. Rearranging gives:
\[ R = \frac{v^2}{a_c} \]
Substituting the values we have:
\[ R = \frac{(10)^2}{5} = \frac{100}{5} = 20 \, \text{m} \]

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

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