A projectile is projected at `10ms^(-1)` by making an angle `60^(@)` to the horizontal. After sometime, its velocity makes an angle of `30^(@)` to the horzontal . Its speed at this instant is:

Options

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

Explanation

To find the speed of the projectile when its velocity makes an angle of \(30^\circ\) with the horizontal, we can follow these steps:

### Step 1: Identify the Initial Conditions
The projectile is projected with an initial speed of \(10 \, \text{m/s}\) at an angle of \(60^\circ\) to the horizontal.

### Step 2: Calculate the Horizontal Component of the Initial Velocity
The horizontal component of the initial velocity (\(u_x\)) can be calculated using the cosine of the angle:
\[ u_x = u \cdot \cos(60^\circ) = 10 \cdot \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

### Step 3: Understand the Horizontal Velocity in Projectile Motion
In projectile motion, the horizontal component of velocity remains constant throughout the motion. Therefore, at any point in time, the horizontal component of the velocity (\(V_x\)) will also be \(5 \, \text{m/s}\).

### Step 4: Set Up the Equation for the Horizontal Component at the Instant of Interest
At the instant when the velocity makes an angle of \(30^\circ\) with the horizontal, we can express the horizontal component of the velocity (\(V_x\)) as:
\[ V_x = V \cdot \cos(30^\circ) \]
Where \(V\) is the speed at that instant.

### Step 5: Substitute the Known Values
Since we know \(V_x = 5 \, \text{m/s}\), we can set up the equation:
\[ 5 = V \cdot \cos(30^\circ) \]
Using \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we can rewrite the equation:
\[ 5 = V \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Solve for \(V\)
Now, we can solve for \(V\):
\[ V = \frac{5 \cdot 2}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s} \]

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