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 find the maximum height attained by a projectile projected at an angle of \(60^\circ\) with the horizontal, we can follow these steps:

### Step 1: Understand the initial conditions
The projectile is launched at an angle of \(60^\circ\) with an initial velocity \(v\). After \(1\) second, the projectile makes an angle of \(30^\circ\) with the horizontal.

### Step 2: Determine the components of the initial velocity
The initial velocity can be broken down into horizontal and vertical components:
- \(v_x = v \cos(60^\circ) = \frac{v}{2}\)
- \(v_y = v \sin(60^\circ) = \frac{\sqrt{3}}{2}v\)

### Step 3: Calculate the vertical velocity after 1 second
The vertical velocity at any time \(t\) is given by:
\[ v_y(t) = v_y - g t \]
Substituting \(t = 1\) second and \(g = 10 \, \text{m/s}^2\):
\[ v_y(1) = \frac{\sqrt{3}}{2}v - 10 \]

### Step 4: Use the angle after 1 second to find the relationship
At \(t = 1\) second, the projectile makes an angle of \(30^\circ\) with the horizontal. Therefore, we can use the tangent of the angle:
\[ \tan(30^\circ) = \frac{v_y(1)}{v_x} \]
Substituting the known values:
\[ \frac{1}{\sqrt{3}} = \frac{\frac{\sqrt{3}}{2}v - 10}{\frac{v}{2}} \]

### Step 5: Solve for \(v\)
Cross multiplying gives:
\[ \frac{v}{2} = \sqrt{3} \left(\frac{\sqrt{3}}{2}v - 10\right) \]
Expanding and simplifying:
\[ \frac{v}{2} = \frac{3}{2}v - 10\sqrt{3} \]
Multiplying through by \(2\) to eliminate the fraction:
\[ v = 3v - 20\sqrt{3} \]
Rearranging gives:
\[ 2v = 20\sqrt{3} \quad \Rightarrow \quad v = 10\sqrt{3} \]

### Step 6: Calculate the maximum height
The formula for maximum height \(H\) is given by:
\[ H = \frac{v^2 \sin^2(\theta)}{2g} \]
Substituting \(v = 10\sqrt{3}\), \(\theta = 60^\circ\), and \(g = 10 \, \text{m/s}^2\):
\[ H = \frac{(10\sqrt{3})^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2 \cdot 10} \]
Calculating:
\[ H = \frac{300 \cdot \frac{3}{4}}{20} = \frac{225}{20} = 11.25 \, \text{m} \]

### Step 7: Final height calculation
To convert to a more standard form:
\[ H = 11.25 \, \text{m} = \frac{45}{4} \, \text{m} \]

### Conclusion
The maximum height attained by the projectile is \( \frac{45}{4} \, \text{m} \).