A grass hopper can jump maximum distance `1.6m.` It spends negligible time on ground. How far can it go in `10(sqrt2)`s?

Options

A. 45 m
B. 30 m
C. 20 m
D. 40 m

Explanation

To solve the problem of how far a grasshopper can jump in \(10\sqrt{2}\) seconds given that it can jump a maximum distance of \(1.6\) meters, we can follow these steps:

### Step 1: Understand the Maximum Range
The maximum range \(R_{\text{max}}\) of a projectile is given by the formula:
\[ R_{\text{max}} = \frac{u^2}{g} \]
where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

### Step 2: Calculate Initial Velocity
Given that \(R_{\text{max}} = 1.6 \, \text{m}\) and \(g = 10 \, \text{m/s}^2\), we can rearrange the formula to find \(u\):
\[ u^2 = R_{\text{max}} \cdot g = 1.6 \cdot 10 = 16 \]
Taking the square root gives:
\[ u = \sqrt{16} = 4 \, \text{m/s} \]

### Step 3: Determine Horizontal Component of Velocity
The horizontal component of the velocity when the angle of projection \(\theta\) is \(45^\circ\) is given by:
\[ u_x = u \cos \theta = 4 \cos 45^\circ \]
Since \(\cos 45^\circ = \frac{1}{\sqrt{2}}\):
\[ u_x = 4 \cdot \frac{1}{\sqrt{2}} = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \, \text{m/s} \]

### Step 4: Calculate Total Distance Covered
The total distance \(s\) covered by the grasshopper in time \(t = 10\sqrt{2}\) seconds is given by:
\[ s = u_x \cdot t = (2\sqrt{2}) \cdot (10\sqrt{2}) \]
Calculating this gives:
\[ s = 2 \cdot 10 \cdot (\sqrt{2} \cdot \sqrt{2}) = 20 \cdot 2 = 40 \, \text{m} \]

### Final Answer
Thus, the total distance the grasshopper can jump in \(10\sqrt{2}\) seconds is:
\[ \boxed{40 \, \text{m}} \]