A grass hopper can jump a maximum horizontal distance of 0.3 m. If it spends negligible time on the ground, its horizontal component of velocity is `(g=10 m//s^(2))`

Options

A. 3/2 m/s
B. `sqrt((3)/(2)) m//s`
C. 1/2 m/s
D. `sqrt((2)/(3)) m//s`

Explanation

To find the horizontal component of the velocity of the grasshopper, we can follow these steps:

### Step 1: Understand the Range Formula
The range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( u \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Identify Maximum Range Condition
The maximum range occurs when \( \sin(2\theta) = 1 \). This happens when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \). Therefore, the formula simplifies to:
\[ R = \frac{u^2}{g} \]

### Step 3: Substitute Known Values
Given that the maximum horizontal distance \( R = 0.3 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \), we can substitute these values into the simplified range formula:
\[ 0.3 = \frac{u^2}{10} \]

### Step 4: Solve for \( u^2 \)
Rearranging the equation to solve for \( u^2 \):
\[ u^2 = 0.3 \times 10 = 3 \]

### Step 5: Calculate \( u \)
Taking the square root of both sides to find \( u \):
\[ u = \sqrt{3} \, \text{m/s} \]

### Step 6: Find the Horizontal Component of Velocity
Since the angle of projection is \( 45^\circ \), the horizontal component of the velocity \( u_x \) can be calculated as:
\[ u_x = u \cos(45^\circ) = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \, \text{m/s} \]

### Final Answer
Thus, the horizontal component of the velocity of the grasshopper is:
\[ u_x = \frac{\sqrt{6}}{2} \, \text{m/s} \]