A particle is projected at an angle of `60^(@)` above the horizontal with a speed of `10m//s`. After some time the direction of its velocity makes an angle of `30^(@)` above the horizontal. The speed of the particle at this instant is s

Options

A. `5/(sqrt(3))m//s`
B. `5sqrt(3)m//s`
C. `5m//s`
D. `10/(sqrt(3))m//s`

Answer

To solve the problem step by step, we will analyze the motion of the particle projected at an angle of \(60^\circ\) with an initial speed of \(10 \, \text{m/s}\) and find the speed when the direction of its velocity makes an angle of \(30^\circ\) above the horizontal.

### Step 1: Determine the initial velocity components
The initial velocity \(u\) can be broken down into its horizontal and vertical components using trigonometric functions.

- The horizontal component \(u_x\) is given by:
\[ u_x = u \cdot \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

- The vertical component \(u_y\) is given by:
\[ u_y = u \cdot \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 2: Analyze the horizontal motion
The horizontal velocity \(v_x\) remains constant throughout the projectile motion since there is no horizontal acceleration (assuming no air resistance):
\[ v_x = u_x = 5 \, \text{m/s} \]

### Step 3: Analyze the vertical motion
The vertical component of the velocity \(v_y\) changes due to the acceleration due to gravity \(g\). The vertical velocity at any time \(t\) can be expressed as:
\[ v_y = u_y - g \cdot t \]

### Step 4: Use the angle of the velocity at the instant
At the instant when the direction of the velocity makes an angle of \(30^\circ\) above the horizontal, we can relate the horizontal and vertical components of the velocity using the tangent function:
\[ \tan(30^\circ) = \frac{v_y}{v_x} \]
From trigonometric values, we know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus, we can write:
\[ \frac{v_y}{5} = \frac{1}{\sqrt{3}} \implies v_y = \frac{5}{\sqrt{3}} \, \text{m/s} \]

### Step 5: Calculate the speed of the particle
The total speed \(s\) of the particle can be calculated using the Pythagorean theorem:
\[ s = \sqrt{v_x^2 + v_y^2} \]
Substituting the values we have:
\[ s = \sqrt{(5)^2 + \left(\frac{5}{\sqrt{3}}\right)^2} \]
Calculating further:
\[ s = \sqrt{25 + \frac{25}{3}} = \sqrt{25 + 8.33} = \sqrt{33.33} = \frac{5\sqrt{3}}{3} \, \text{m/s} \]

### Final Answer
Thus, the speed of the particle at the instant when its velocity makes an angle of \(30^\circ\) above the horizontal is:
\[ s = \frac{10}{\sqrt{3}} \, \text{m/s} \]