A heavy particle is projected from a point on the horizontal at an angle `45^(@)` wil the horizontal with a speed of 20 m/s. Then the radius of the curvature of its path at the instant of crossing the same horizontal is ______

Options

A. `10 sqrt(2)`
B. `40 sqrt(2)`
C. `20 sqrt(2)`
D. none of these

Explanation

To find the radius of curvature of the path of a heavy particle projected at an angle of \(45^\circ\) with a speed of \(20 \, \text{m/s}\) at the instant it crosses the horizontal, we can follow these steps:

### Step 1: Identify the initial conditions
The particle is projected with:
- Initial speed \(u = 20 \, \text{m/s}\)
- Angle of projection \(\theta = 45^\circ\)

### Step 2: Determine the velocity components
The horizontal and vertical components of the initial velocity can be calculated as:
- \(u_x = u \cos \theta = 20 \cos 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s}\)
- \(u_y = u \sin \theta = 20 \sin 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s}\)

### Step 3: Calculate the velocity when crossing the horizontal
At the instant of crossing the horizontal again, the vertical component of the velocity will be equal to the initial vertical component (due to symmetry in projectile motion). Therefore, the total velocity \(v\) at this point will be:
- \(v = \sqrt{u_x^2 + v_y^2}\)
- Since \(v_y\) at this point is also \(10\sqrt{2} \, \text{m/s}\) (the same as \(u_y\)), we have:
- \(v = \sqrt{(10\sqrt{2})^2 + (10\sqrt{2})^2} = \sqrt{200 + 200} = \sqrt{400} = 20 \, \text{m/s}\)

### Step 4: Determine the acceleration components
The only acceleration acting on the particle is due to gravity, which acts downwards. The acceleration perpendicular to the velocity at the point of crossing the horizontal is given by:
- \(a_{\perp} = g \cos \theta = g \cos 45^\circ = g \times \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}\)

Using \(g \approx 10 \, \text{m/s}^2\):
- \(a_{\perp} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s}^2\)

### Step 5: Calculate the radius of curvature
The radius of curvature \(R\) can be calculated using the formula:
\[ R = \frac{v^2}{a_{\perp}} \]
Substituting the values we have:
\[ R = \frac{(20)^2}{5\sqrt{2}} = \frac{400}{5\sqrt{2}} = \frac{80}{\sqrt{2}} = 40\sqrt{2} \, \text{m} \]

### Final Answer
The radius of curvature of the particle's path at the instant of crossing the horizontal is \(40\sqrt{2} \, \text{m}\).

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