A projectile is projected from the ground by making an angle of `60^@` with the horizontal. After 1 s projectile makes an angle of `30^@` with the horizontal . The maximum height attained by the projectile is (Take `g=10 ms^-2)`

Options

A. `45/2m`
B. `45/4m`
C. `43/2m`
D. `43/4m`

Answer

To solve the problem step by step, we need to determine the maximum height attained by a projectile projected at an angle of \(60^\circ\) with respect to the horizontal. We know that after 1 second, the projectile makes an angle of \(30^\circ\) with the horizontal. We will use the equations of motion and some trigonometric relationships to find the solution.

### Step 1: Identify the components of the initial velocity
The initial velocity \(u\) can be broken down into its horizontal and vertical components:
- Horizontal component: \(u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2}\)
- Vertical component: \(u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2}\)

### Step 2: Determine the vertical velocity after 1 second
The vertical velocity \(V_y\) after 1 second can be calculated using the equation of motion:
\[ V_y = u_y - g \cdot t \]
Substituting the values, we have:
\[ V_y = u \cdot \frac{\sqrt{3}}{2} - 10 \cdot 1 \]
\[ V_y = u \cdot \frac{\sqrt{3}}{2} - 10 \]

### Step 3: Relate the vertical and horizontal velocities at \(30^\circ\)
At \(t = 1\) second, the projectile makes an angle of \(30^\circ\) with the horizontal. Thus, we can use the tangent of the angle:
\[ \tan(30^\circ) = \frac{V_y}{V_x} \]
Where \(V_x\) is the horizontal component of the velocity, which remains constant:
\[ V_x = u_x = \frac{u}{2} \]
Substituting the values:
\[ \frac{1}{\sqrt{3}} = \frac{u \cdot \frac{\sqrt{3}}{2} - 10}{\frac{u}{2}} \]

### Step 4: Simplify the equation
Cross-multiplying gives:
\[ \frac{u}{2} = \sqrt{3} \left(u \cdot \frac{\sqrt{3}}{2} - 10\right) \]
Expanding and simplifying:
\[ \frac{u}{2} = \frac{3u}{2} - 10\sqrt{3} \]
Rearranging gives:
\[ 10\sqrt{3} = \frac{3u}{2} - \frac{u}{2} \]
\[ 10\sqrt{3} = \frac{2u}{2} = u \]
Thus, we find:
\[ u = 20\sqrt{3} \text{ m/s} \]

### Step 5: Calculate the maximum height
The maximum height \(h\) attained by the projectile can be calculated using the formula:
\[ h = \frac{u^2 \sin^2(60^\circ)}{2g} \]
Substituting the values:
\[ h = \frac{(20\sqrt{3})^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2 \cdot 10} \]
Calculating:
\[ h = \frac{1200 \cdot \frac{3}{4}}{20} \]
\[ h = \frac{900}{20} = 45 \text{ m} \]

### Final Answer
The maximum height attained by the projectile is \( \frac{45}{4} \) meters.