A grasshopper can jump a maximum horizontal distance of 40cm. If it spends negligible time on the ground then in this case its speed along the horizontal raod will be

Options

A. 1 m/s
B. 2 m/s
C. `sqrt2 m//s`
D. `2sqrt2 m//s`

Answer

To solve the problem step by step, we will analyze the motion of the grasshopper and apply the relevant physics concepts.

### Step 1: Understand the Problem
The grasshopper can jump a maximum horizontal distance (range) of 40 cm. We need to find its speed along the horizontal road when it spends negligible time on the ground.

### Step 2: Convert Units
Convert the range from centimeters to meters for consistency in SI units:
\[ R = 40 \text{ cm} = \frac{40}{100} = 0.4 \text{ m} \]

### Step 3: Use the Range Formula
The formula for the range \( R \) of a projectile launched at an angle \( \theta \) with an initial speed \( u \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 4: Determine the Angle for Maximum Range
The maximum range for projectile motion occurs at an angle of \( 45^\circ \). Therefore, we set \( \theta = 45^\circ \):
\[ \sin(2\theta) = \sin(90^\circ) = 1 \]

### Step 5: Substitute Values into the Range Formula
Now, substituting the known values into the range formula:
\[ 0.4 = \frac{u^2 \cdot 1}{10} \]

### Step 6: Solve for Initial Speed \( u \)
Rearranging the equation to solve for \( u^2 \):
\[ u^2 = 0.4 \times 10 = 4 \]
Taking the square root gives:
\[ u = \sqrt{4} = 2 \, \text{m/s} \]

### Step 7: Find the Horizontal Component of Velocity
The horizontal component of the initial velocity \( u_x \) when the angle is \( 45^\circ \) is given by:
\[ u_x = u \cos(45^\circ) \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ u_x = 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{m/s} \]

### Final Answer
The speed along the horizontal road is:
\[ \sqrt{2} \, \text{m/s} \]