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

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 projectile motion of the particle and use the given information to find the speed of the particle when its velocity makes a 30-degree angle with the horizontal.

### Step 1: Identify the initial conditions
The particle is projected at an angle of \(60^\circ\) with an initial speed of \(10 \, \text{m/s}\). We can break this initial velocity into its horizontal and vertical components.

- Horizontal component (\(u_x\)):
\[ u_x = u \cos(60^\circ) = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

- Vertical component (\(u_y\)):
\[ u_y = u \sin(60^\circ) = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 2: Analyze the motion at the instant when the angle is \(30^\circ\)
At the instant when the velocity makes an angle of \(30^\circ\) above the horizontal, we can denote the speed of the particle at that moment as \(V\). The horizontal and vertical components of the velocity at this instant can be expressed as:

- Horizontal component (\(V_x\)):
\[ V_x = V \cos(30^\circ) = V \times \frac{\sqrt{3}}{2} \]

- Vertical component (\(V_y\)):
\[ V_y = V \sin(30^\circ) = V \times \frac{1}{2} \]

### Step 3: Use the conservation of horizontal velocity
In projectile motion, the horizontal component of velocity remains constant. Therefore, we can set the horizontal component of the initial velocity equal to the horizontal component of the velocity at the instant when the angle is \(30^\circ\):

\[ u_x = V_x \]
\[ 5 \, \text{m/s} = V \cos(30^\circ) \]
Substituting \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ 5 = V \times \frac{\sqrt{3}}{2} \]

### Step 4: Solve for \(V\)
Now we can solve for \(V\):
\[ V = \frac{5 \times 2}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s} \]

### Step 5: Conclusion
The speed of the particle at the instant when the direction of its velocity makes an angle of \(30^\circ\) above the horizontal is:
\[ \boxed{\frac{10}{\sqrt{3}} \, \text{m/s}} \]