A ball thrown in air follows a path given by `y= x/sqrt3 -(3g)/8 x^2 m` where x-axis is taken along the horizontal and y-axis along the vertical. The maximum displacement of the ball along x-direction for which displacement along y is zero equals to

Options

A. `15 sqrt3m`
B. `(4//15 sqrt3)m`
C. `4//3 m`
D. data insufficient

Explanation

To solve the problem, we need to find the maximum displacement along the x-direction (denoted as \( x \)) for which the displacement along the y-direction (denoted as \( y \)) is zero. The path of the ball is given by the equation:

\[ y = \frac{x}{\sqrt{3}} - \frac{3g}{8} x^2 \]

### Step 1: Set the y-equation to zero

To find the points where the displacement along the y-direction is zero, we set \( y \) to zero:

\[ 0 = \frac{x}{\sqrt{3}} - \frac{3g}{8} x^2 \]

### Step 2: Factor out x

Rearranging the equation gives us:

\[ \frac{x}{\sqrt{3}} = \frac{3g}{8} x^2 \]

Now, we can factor out \( x \):

\[ x \left( \frac{1}{\sqrt{3}} - \frac{3g}{8} x \right) = 0 \]

### Step 3: Solve for x

This gives us two solutions:

1. \( x = 0 \)
2. \( \frac{1}{\sqrt{3}} - \frac{3g}{8} x = 0 \)

From the second equation, we can solve for \( x \):

\[ \frac{3g}{8} x = \frac{1}{\sqrt{3}} \]

\[ x = \frac{8}{3g} \cdot \frac{1}{\sqrt{3}} = \frac{8}{3\sqrt{3}g} \]

### Step 4: Substitute the value of g

Assuming \( g = 10 \, \text{m/s}^2 \):

\[ x = \frac{8}{3\sqrt{3} \cdot 10} = \frac{8}{30\sqrt{3}} = \frac{4}{15\sqrt{3}} \, \text{m} \]

### Conclusion

Thus, the maximum displacement of the ball along the x-direction for which the displacement along the y-direction is zero is:

\[ \boxed{\frac{4}{15\sqrt{3}} \, \text{m}} \]