If a missile is propelled horizontally with 20 m/s from some height, what is the value of `R_(C)` at t=2s?

Answer

To find the radius of curvature \( R_C \) of a missile propelled horizontally with a speed of 20 m/s from a height after 2 seconds, we can follow these steps:

### Step 1: Determine the horizontal and vertical velocities

The horizontal velocity \( v_x \) remains constant:
\[ v_x = 20 \, \text{m/s} \]

For the vertical velocity \( v_y \) after time \( t = 2 \, \text{s} \), we use the equation of motion:
\[ v_y = u_y + g \cdot t \]
where:
- \( u_y = 0 \) (initial vertical velocity)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 2 \, \text{s} \)

Substituting the values:
\[ v_y = 0 + 10 \cdot 2 = 20 \, \text{m/s} \]

### Step 2: Calculate the net velocity

The net velocity \( v_{net} \) is the vector sum of the horizontal and vertical velocities:
\[ v_{net} = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v_{net} = \sqrt{(20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s} \]

### Step 3: Determine the acceleration component

The only acceleration acting on the missile is due to gravity, which acts vertically downwards. The radius of curvature \( R_C \) is given by:
\[ R_C = \frac{v_{net}^2}{a} \]
where \( a \) is the component of acceleration perpendicular to the velocity vector. Since the vertical and horizontal components of velocity are equal, the angle between them is \( 45^\circ \). The component of gravity acting perpendicular to the net velocity is:
\[ a = g \cdot \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s}^2 \]

### Step 4: Calculate the radius of curvature

Substituting the values into the formula for \( R_C \):
\[ R_C = \frac{(20\sqrt{2})^2}{5\sqrt{2}} = \frac{800}{5\sqrt{2}} = \frac{160}{\sqrt{2}} = 80\sqrt{2} \, \text{m} \]

### Final Answer
The radius of curvature \( R_C \) at \( t = 2 \, \text{s} \) is:
\[ R_C = 80\sqrt{2} \, \text{m} \]

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