A projectile is thrown with a velocity of 18 m/s at an angle of `60^@` with horizontal. The interval between the moment when speed is 15 m/s is `(g = 10 m/s^2)`

Options

A. `(12/5)S`
B. `(4/5)S`
C. 12 S
D. `(2/5)S`

Answer

To solve the problem, we need to find the time interval during which the speed of the projectile is 15 m/s. The projectile is thrown with an initial velocity of 18 m/s at an angle of 60 degrees with the horizontal. We will break down the steps systematically.

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

- Horizontal component \( u_x = u \cos(\theta) \)
- Vertical component \( u_y = u \sin(\theta) \)

Given:
- \( u = 18 \, \text{m/s} \)
- \( \theta = 60^\circ \)

Calculating the components:
\[ u_x = 18 \cos(60^\circ) = 18 \times \frac{1}{2} = 9 \, \text{m/s} \]
\[ u_y = 18 \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \, \text{m/s} \]

### Step 2: Write the expression for speed
The speed \( v \) of the projectile at any time \( t \) can be expressed as:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Where \( v_x \) is constant and equal to \( u_x \), and \( v_y \) changes due to gravity.

### Step 3: Set up the equation for speed
We need to find the time when the speed \( v = 15 \, \text{m/s} \):
\[ 15 = \sqrt{u_x^2 + v_y^2} \]
Substituting \( u_x = 9 \, \text{m/s} \):
\[ 15 = \sqrt{9^2 + v_y^2} \]
Squaring both sides:
\[ 225 = 81 + v_y^2 \]
\[ v_y^2 = 225 - 81 = 144 \]
\[ v_y = \pm 12 \, \text{m/s} \]

### Step 4: Determine the time for each case of \( v_y \)
Using the vertical motion equation:
\[ v_y = u_y - g t \]
For \( v_y = 12 \, \text{m/s} \):
\[ 12 = 9\sqrt{3} - 10t_1 \]
Rearranging gives:
\[ 10t_1 = 9\sqrt{3} - 12 \]
\[ t_1 = \frac{9\sqrt{3} - 12}{10} \]

For \( v_y = -12 \, \text{m/s} \):
\[ -12 = 9\sqrt{3} - 10t_2 \]
Rearranging gives:
\[ 10t_2 = 9\sqrt{3} + 12 \]
\[ t_2 = \frac{9\sqrt{3} + 12}{10} \]

### Step 5: Calculate the time interval
The time interval \( \Delta t \) is given by:
\[ \Delta t = t_2 - t_1 \]
Substituting the values:
\[ \Delta t = \left(\frac{9\sqrt{3} + 12}{10}\right) - \left(\frac{9\sqrt{3} - 12}{10}\right) \]
\[ \Delta t = \frac{(9\sqrt{3} + 12) - (9\sqrt{3} - 12)}{10} \]
\[ \Delta t = \frac{24}{10} = 2.4 \, \text{s} \]

### Final Answer
The time interval during which the speed of the projectile is 15 m/s is **2.4 seconds**.