A projectile is thrown with a velocity of `10 ms^(-1)` at an angle of `60^(@)` with horizontal. The interval between the moments when speed is `sqrt(5g) m//s` is (Take, `g = 10 ms^(-2))`

Options

A. 1s
B. 3s
C. 2s
D. 4s

Explanation

To solve the problem of finding the time interval between the moments when the speed of a projectile is \( \sqrt{5g} \, \text{m/s} \), we will follow these steps:

### Step 1: Identify the given parameters
- Initial velocity \( u = 10 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
- The horizontal component \( V_x \) is given by:
\[ V_x = u \cos \theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]
- The vertical component \( V_y \) can be expressed as:
\[ V_y = u \sin \theta = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 3: Write the expression for the vertical velocity at time \( t \)
The vertical velocity \( V_y(t) \) at time \( t \) is given by:
\[ V_y(t) = V_{y0} - gt = 5\sqrt{3} - 10t \]

### Step 4: Calculate the resultant velocity
The resultant velocity \( V \) at any time \( t \) is given by:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V = \sqrt{(5)^2 + (5\sqrt{3} - 10t)^2} \]
Setting this equal to \( \sqrt{5g} \):
\[ \sqrt{5g} = \sqrt{5 \times 10} = \sqrt{50} = 5\sqrt{2} \]
Thus, we have:
\[ \sqrt{(5)^2 + (5\sqrt{3} - 10t)^2} = 5\sqrt{2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (5)^2 + (5\sqrt{3} - 10t)^2 = (5\sqrt{2})^2 \]
\[ 25 + (5\sqrt{3} - 10t)^2 = 50 \]
This simplifies to:
\[ (5\sqrt{3} - 10t)^2 = 25 \]

### Step 6: Solve for \( t \)
Taking the square root of both sides:
\[ 5\sqrt{3} - 10t = 5 \quad \text{or} \quad 5\sqrt{3} - 10t = -5 \]
1. For \( 5\sqrt{3} - 10t = 5 \):
\[ 10t = 5\sqrt{3} - 5 \implies t_1 = \frac{5(\sqrt{3} - 1)}{10} = \frac{\sqrt{3} - 1}{2} \]
2. For \( 5\sqrt{3} - 10t = -5 \):
\[ 10t = 5\sqrt{3} + 5 \implies t_2 = \frac{5(\sqrt{3} + 1)}{10} = \frac{\sqrt{3} + 1}{2} \]

### Step 7: Calculate the time interval
The time interval \( \Delta t \) between these two moments is:
\[ \Delta t = t_2 - t_1 = \left(\frac{\sqrt{3} + 1}{2}\right) - \left(\frac{\sqrt{3} - 1}{2}\right) \]
This simplifies to:
\[ \Delta t = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{2} = \frac{2}{2} = 1 \, \text{s} \]

### Final Answer
The interval between the moments when the speed is \( \sqrt{5g} \, \text{m/s} \) is **1 second**.

---