A particle is projected from horizontal making an angle of `53^(@)` with initial velocity `100 ms^(-1)`. The time taken by the particle to make angle `45^(@)` from horizontal is

Options

A. 14s
B. 2s
C. Both (a) and (b)
D. None of these

Explanation

To solve the problem step by step, we will analyze the projectile motion of the particle projected at an angle of \(53^\circ\) with an initial velocity of \(100 \, \text{m/s}\) and determine the time taken for it to make an angle of \(45^\circ\) with the horizontal.

### Step 1: Determine the Components of the Initial Velocity
The initial velocity can be broken down into horizontal and vertical components using trigonometric functions.

- **Horizontal Component (\(u_x\))**:
\[ u_x = u \cos \theta = 100 \cos(53^\circ) \]
Using \(\cos(53^\circ) \approx 0.6\):
\[ u_x = 100 \times 0.6 = 60 \, \text{m/s} \]

- **Vertical Component (\(u_y\))**:
\[ u_y = u \sin \theta = 100 \sin(53^\circ) \]
Using \(\sin(53^\circ) \approx 0.8\):
\[ u_y = 100 \times 0.8 = 80 \, \text{m/s} \]

### Step 2: Analyze the Condition for \(45^\circ\) Angle
For the particle to make an angle of \(45^\circ\) with the horizontal, the vertical and horizontal components of the velocity must be equal at that point in time.

Let \(v_y\) be the vertical component of the velocity when the angle is \(45^\circ\). Therefore:
\[ v_y = u_x = 60 \, \text{m/s} \]

### Step 3: Use the Equation of Motion to Find Time
We can use the equation of motion to find the time when the vertical component of the velocity becomes \(60 \, \text{m/s}\):
\[ v_y = u_y - g t \]
Where:
- \(g\) is the acceleration due to gravity (\(10 \, \text{m/s}^2\)),
- \(t\) is the time.

Substituting the known values:
\[ 60 = 80 - 10t \]

### Step 4: Solve for Time \(t\)
Rearranging the equation:
\[ 10t = 80 - 60 \]
\[ 10t = 20 \]
\[ t = 2 \, \text{seconds} \]

### Step 5: Check for the Second Instance
The particle will also make a \(45^\circ\) angle again when it is descending. We can find this time as follows:

Using the same equation:
\[ v_y = u_y - g t \]
At the second instance, the vertical velocity will be \(-60 \, \text{m/s}\) (downward):
\[ -60 = 80 - 10t \]

Rearranging gives:
\[ 10t = 80 + 60 \]
\[ 10t = 140 \]
\[ t = 14 \, \text{seconds} \]

### Conclusion
The times at which the particle makes a \(45^\circ\) angle with the horizontal are \(2 \, \text{seconds}\) and \(14 \, \text{seconds}\).