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 of the ball along the x-direction for which the displacement along the y-direction is zero. The trajectory of the ball is given by the equation:

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

### Step 1: Set the y-coordinate to zero
To find the maximum displacement along the x-direction where the y-displacement is zero, we set \( y = 0 \):

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

### Step 2: Rearrange the equation
Rearranging the equation gives us:

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

### Step 3: Factor out x
We can factor out \( x \) from the equation:

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

This gives us two solutions:

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

### Step 4: Solve for x
Now, we solve for \( x \) from the second equation:

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

Multiplying both sides by \( \frac{8}{3g} \):

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

### Step 5: Substitute the value of g
Assuming \( g = 10 \, \text{m/s}^2 \):

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

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

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

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