A body is thrown with the velocity `20ms^(-1)` at an angle of `60^@` with the horizontal. Find the time gap between the two positions of the body where the velocity of the body makes an angle of `30^@` with horizontal.

Options

A. 1.15 s
B. 0.95 s
C. 1 s
D. 1.5 s

Answer

To solve the problem, we need to find the time gap between two positions of a body thrown at an angle, where the velocity makes an angle of \(30^\circ\) with the horizontal. Here’s a step-by-step solution:

### Step 1: Break down the initial velocity into components
The body is thrown with an initial velocity \(u = 20 \, \text{m/s}\) at an angle of \(60^\circ\).

- **Horizontal component** (\(u_x\)):
\[ u_x = u \cos(60^\circ) = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

- **Vertical component** (\(u_y\)):
\[ u_y = u \sin(60^\circ) = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 2: Determine the conditions for the velocity to be at \(30^\circ\)
At the two instances when the velocity makes an angle of \(30^\circ\) with the horizontal, we can use the tangent of the angle to relate the vertical and horizontal components of the velocity.

Using the tangent relationship:
\[ \tan(30^\circ) = \frac{v_y}{v_x} \]
Where \(v_x\) remains constant at \(10 \, \text{m/s}\). Thus:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \Rightarrow v_y = 10 \cdot \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s} \]

### Step 3: Set up the equations for the two instances
1. **First instance** (upward motion):
\[ v_y = u_y - g t_1 = 10\sqrt{3} - 9.8 t_1 \]
Setting \(v_y = \frac{10}{\sqrt{3}}\):
\[ \frac{10}{\sqrt{3}} = 10\sqrt{3} - 9.8 t_1 \]
Rearranging gives:
\[ 9.8 t_1 = 10\sqrt{3} - \frac{10}{\sqrt{3}} \]

2. **Second instance** (downward motion):
\[ v_y = u_y - g t_2 = 10\sqrt{3} - 9.8 t_2 \]
Setting \(v_y = -\frac{10}{\sqrt{3}}\):
\[ -\frac{10}{\sqrt{3}} = 10\sqrt{3} - 9.8 t_2 \]
Rearranging gives:
\[ 9.8 t_2 = 10\sqrt{3} + \frac{10}{\sqrt{3}} \]

### Step 4: Solve for \(t_1\) and \(t_2\)
1. For \(t_1\):
\[ 9.8 t_1 = 10\sqrt{3} - \frac{10}{\sqrt{3}} = 10\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 10\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{20}{\sqrt{3}} \]
\[ t_1 = \frac{20}{9.8\sqrt{3}} \]

2. For \(t_2\):
\[ 9.8 t_2 = 10\sqrt{3} + \frac{10}{\sqrt{3}} = 10\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) = 10\left(\frac{3 + 1}{\sqrt{3}}\right) = \frac{40}{\sqrt{3}} \]
\[ t_2 = \frac{40}{9.8\sqrt{3}} \]

### Step 5: Find the time gap \(t_2 - t_1\)
\[ t_2 - t_1 = \frac{40}{9.8\sqrt{3}} - \frac{20}{9.8\sqrt{3}} = \frac{20}{9.8\sqrt{3}} \]
Calculating this gives:
\[ t_2 - t_1 \approx 1.17 \, \text{s} \]

### Final Answer:
The time gap between the two positions of the body where the velocity makes an angle of \(30^\circ\) with the horizontal is approximately \(1.17 \, \text{s}\).