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 of finding the time gap between two positions of a body thrown at a velocity of \(20 \, \text{m/s}\) at an angle of \(60^\circ\) with the horizontal, where the velocity makes an angle of \(30^\circ\) with the horizontal, we can follow these steps:

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

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

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

### Step 2: Use the conditions for the velocity making an angle of \(30^\circ\)
At the two positions where the velocity makes an angle of \(30^\circ\) with the horizontal, we can analyze the velocity components at these points.

Let \(V\) be the velocity at these points. The horizontal component remains constant:
\[ V_x = U_x = 10 \, \text{m/s} \]

The vertical component \(V_y\) can be related to the angle:
\[ \tan(30^\circ) = \frac{V_y}{V_x} \Rightarrow V_y = V_x \tan(30^\circ) = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s} \]

### Step 3: Apply the equations of motion
Using the vertical motion equation, we can find the time at which the body reaches this vertical velocity component. The vertical velocity at any time \(t\) is given by:
\[ V_y = U_y - g t \]
Substituting the known values:
\[ \frac{10}{\sqrt{3}} = 10\sqrt{3} - 9.8 t \]

### Step 4: Solve for time \(t\)
Rearranging the equation:
\[ 9.8 t = 10\sqrt{3} - \frac{10}{\sqrt{3}} \]
Calculating \(10\sqrt{3} - \frac{10}{\sqrt{3}}\):
\[ 10\sqrt{3} - \frac{10}{\sqrt{3}} = 10\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 10\left(\frac{3 - 1}{\sqrt{3}}\right) = 10 \times \frac{2}{\sqrt{3}} = \frac{20}{\sqrt{3}} \]

Now substituting back:
\[ 9.8 t = \frac{20}{\sqrt{3}} \Rightarrow t = \frac{20}{9.8\sqrt{3}} \]

### Step 5: Calculate the total time gap
Since the body will reach the angle of \(30^\circ\) twice (once while going up and once while coming down), the time gap between these two positions is:
\[ \Delta t = 2t = 2 \times \frac{20}{9.8\sqrt{3}} = \frac{40}{9.8\sqrt{3}} \approx \frac{40}{16.97} \approx 2.36 \, \text{s} \]

### 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 \(2.36 \, \text{s}\).