A grasshopper can jump a maximum distance `1.6 m`. It spends negligible time on the ground. How far can it go in `10 s` ?

Options

A. `5 sqrt(2) m`
B. `10 sqrt(2) m`
C. `20 sqrt(2) m`
D. `40 sqrt(2) m`

Explanation

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the maximum range of the jump
The maximum distance a grasshopper can jump is given as \( R_{\text{max}} = 1.6 \, \text{m} \). The maximum range for projectile motion occurs at an angle of \( 45^\circ \).

### Step 2: Use the formula for maximum range
The formula for the maximum range \( R \) in projectile motion is given by:
\[ R_{\text{max}} = \frac{u^2 \sin(2\theta)}{g} \]
For \( \theta = 45^\circ \), \( \sin(90^\circ) = 1 \), so the formula simplifies to:
\[ R_{\text{max}} = \frac{u^2}{g} \]

### Step 3: Rearrange to find initial velocity \( u \)
From the maximum range formula, we can rearrange it to find the initial velocity \( u \):
\[ u^2 = R_{\text{max}} \cdot g \]
Substituting \( R_{\text{max}} = 1.6 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \):
\[ u^2 = 1.6 \cdot 10 = 16 \]
Taking the square root:
\[ u = \sqrt{16} = 4 \, \text{m/s} \]

### Step 4: Calculate horizontal component of velocity
The horizontal component of the initial velocity \( u \) when the angle \( \theta = 45^\circ \) is:
\[ u_x = u \cos(45^\circ) = 4 \cdot \frac{1}{\sqrt{2}} = 2\sqrt{2} \, \text{m/s} \]

### Step 5: Calculate total distance covered in 10 seconds
The total distance \( d \) covered in time \( t = 10 \, \text{s} \) is given by:
\[ d = u_x \cdot t = (2\sqrt{2}) \cdot 10 = 20\sqrt{2} \, \text{m} \]

### Final Answer
The grasshopper can cover a distance of \( 20\sqrt{2} \, \text{m} \) in \( 10 \, \text{s} \).

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