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 between the moments when the speed of the projectile is 15 m/s. We will break down the solution step-by-step.

### Step 1: Break the initial velocity into components
The initial velocity \( u = 18 \, \text{m/s} \) is thrown at an angle of \( 60^\circ \) with the horizontal.

- The horizontal component of the velocity:
\[ u_x = u \cos(60^\circ) = 18 \cdot \frac{1}{2} = 9 \, \text{m/s} \]

- The vertical component of the velocity:
\[ u_y = u \sin(60^\circ) = 18 \cdot \frac{\sqrt{3}}{2} = 9\sqrt{3} \, \text{m/s} \]

### Step 2: Write the equations for velocity at time \( t \)
The acceleration in the x-direction is \( 0 \, \text{m/s}^2 \) and in the y-direction is \( -g = -10 \, \text{m/s}^2 \).

The velocity components at time \( t \) are:
- Horizontal:
\[ v_x = u_x = 9 \, \text{m/s} \]

- Vertical:
\[ v_y = u_y - gt = 9\sqrt{3} - 10t \, \text{m/s} \]

### Step 3: Calculate the magnitude of the velocity
The magnitude of the velocity \( v \) at time \( t \) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(9)^2 + (9\sqrt{3} - 10t)^2} \]

### Step 4: Set the magnitude equal to 15 m/s
We want to find when the speed is 15 m/s:
\[ \sqrt{81 + (9\sqrt{3} - 10t)^2} = 15 \]
Squaring both sides:
\[ 81 + (9\sqrt{3} - 10t)^2 = 225 \]
\[ (9\sqrt{3} - 10t)^2 = 144 \]

### Step 5: Solve for \( t \)
Taking the square root of both sides:
\[ 9\sqrt{3} - 10t = \pm 12 \]

#### Case 1:
\[ 9\sqrt{3} - 10t = 12 \]
\[ 10t = 9\sqrt{3} - 12 \]
\[ t = \frac{9\sqrt{3} - 12}{10} \]

#### Case 2:
\[ 9\sqrt{3} - 10t = -12 \]
\[ 10t = 9\sqrt{3} + 12 \]
\[ t = \frac{9\sqrt{3} + 12}{10} \]

### Step 6: Calculate the values of \( t \)
Calculating \( \sqrt{3} \approx 1.732 \):
\[ 9\sqrt{3} \approx 15.588 \]

#### For Case 1:
\[ t_1 = \frac{15.588 - 12}{10} = \frac{3.588}{10} \approx 0.3588 \, \text{s} \]

#### For Case 2:
\[ t_2 = \frac{15.588 + 12}{10} = \frac{27.588}{10} \approx 2.7588 \, \text{s} \]

### Step 7: Find the interval between the two times
The time interval \( \Delta t \) is:
\[ \Delta t = t_2 - t_1 = 2.7588 - 0.3588 \approx 2.4 \, \text{s} \]

### Final Answer
The interval between the moments when the speed is 15 m/s is approximately \( 2.4 \, \text{s} \).

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