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 of finding the time taken by a particle projected at an angle of \(53^\circ\) with an initial velocity of \(100 \, \text{m/s}\) to make an angle of \(45^\circ\) with the horizontal, we can follow these steps:

### Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity \(u\) can be resolved into horizontal (\(u_x\)) and vertical (\(u_y\)) components using trigonometric functions:

\[ u_x = u \cdot \cos(\theta) = 100 \cdot \cos(53^\circ) \]
\[ u_y = u \cdot \sin(\theta) = 100 \cdot \sin(53^\circ) \]

Using the values of \(\cos(53^\circ) = \frac{3}{5}\) and \(\sin(53^\circ) = \frac{4}{5}\):

\[ u_x = 100 \cdot \frac{3}{5} = 60 \, \text{m/s} \]
\[ u_y = 100 \cdot \frac{4}{5} = 80 \, \text{m/s} \]

### Step 2: Determine the conditions for the angle of \(45^\circ\).
At the point where the particle makes an angle of \(45^\circ\) with the horizontal, the horizontal and vertical components of the velocity (\(v_x\) and \(v_y\)) will be equal:

\[ v_x = v_y \]

Since the horizontal component of velocity remains constant:

\[ v_x = u_x = 60 \, \text{m/s} \]

Thus, we have:

\[ v_y = 60 \, \text{m/s} \]

### Step 3: Use the kinematic equation to find the time.
The vertical component of the velocity can be expressed using the kinematic equation:

\[ v_y = u_y + a \cdot t \]

Where:
- \(v_y = 60 \, \text{m/s}\)
- \(u_y = 80 \, \text{m/s}\)
- \(a = -g = -10 \, \text{m/s}^2\) (acceleration due to gravity)

Substituting the values:

\[ 60 = 80 - 10t \]

Rearranging gives:

\[ 10t = 80 - 60 \]
\[ 10t = 20 \]
\[ t = 2 \, \text{s} \]

### Step 4: Find the second time when the angle is \(45^\circ\).
The second instance occurs when the particle is descending. At that point, the vertical velocity will be \(-60 \, \text{m/s}\):

\[ -60 = 80 - 10t \]

Rearranging gives:

\[ -60 - 80 = -10t \]
\[ -140 = -10t \]
\[ t = 14 \, \text{s} \]

### Final Result:
The times at which the particle makes an angle of \(45^\circ\) with the horizontal are \(t_1 = 2 \, \text{s}\) and \(t_2 = 14 \, \text{s}\).