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

### Step 1: Understand the Given Information
We have a projectile thrown with an initial velocity \( v_0 = 10 \, \text{m/s} \) at an angle \( \theta = 60^\circ \) with the horizontal. We need to find the time intervals when the speed is \( \sqrt{5g} \).

### Step 2: Calculate \( \sqrt{5g} \)
Given \( g = 10 \, \text{m/s}^2 \):
\[ \sqrt{5g} = \sqrt{5 \times 10} = \sqrt{50} \, \text{m/s} \]

### Step 3: Determine the Components of Velocity
The horizontal and vertical components of the initial velocity are:
- Horizontal component \( V_x = v_0 \cos \theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)
- Vertical component \( V_y = v_0 \sin \theta - gt = 10 \sin 60^\circ - 10t = 10 \times \frac{\sqrt{3}}{2} - 10t = 5\sqrt{3} - 10t \)

### Step 4: Write the Expression for Speed
The speed \( V \) of the projectile is given by:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the components:
\[ V = \sqrt{(5)^2 + (5\sqrt{3} - 10t)^2} \]
Setting this equal to \( \sqrt{50} \):
\[ \sqrt{50} = \sqrt{25 + (5\sqrt{3} - 10t)^2} \]

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

### Step 6: Solve the Equation
Taking the square root of both sides:
\[ 5 = 5\sqrt{3} - 10t \quad \text{or} \quad 5 = -(5\sqrt{3} - 10t) \]
From the first equation:
\[ 10t = 5\sqrt{3} - 5 \quad \Rightarrow \quad t_1 = \frac{5(\sqrt{3} - 1)}{10} = \frac{\sqrt{3} - 1}{2} \]
From the second equation:
\[ 10t = 5\sqrt{3} + 5 \quad \Rightarrow \quad t_2 = \frac{5(\sqrt{3} + 1)}{10} = \frac{\sqrt{3} + 1}{2} \]

### Step 7: Calculate the Time Interval
The time interval \( |t_2 - t_1| \) is:
\[ |t_2 - t_1| = \left| \frac{\sqrt{3} + 1}{2} - \frac{\sqrt{3} - 1}{2} \right| = \left| \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{2} \right| = \left| \frac{2}{2} \right| = 1 \, \text{s} \]

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

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