A

Step 1: The diameter of the tree trunk is given as 20 cm, which means the radius is \frac{20 \text{ cm}}{2} = 10 \text{ cm} = 0.1 \text{ m}.
Step 2: For the grasshopper to clear the trunk, it must jump a horizontal distance equal to the trunk's diameter. This means it must cover a horizontal distance of 0.2 m.
Step 3: When the grasshopper jumps, it follows a parabolic trajectory. To determine the minimum take-off speed, we use the kinematic equations of motion.
Step 4: Assuming the jump angle is 45° (the optimal angle for maximum distance in projectile motion), we have:
- Horizontal range R = \frac{u^2 \sin(2\theta)}{g} where \theta = 45° and g = 9.81 m/s².
- Hence, R = \frac{u^2}{g}, since \sin(90°) = 1.
Step 5: Setting R equal to 0.2 m gives us \frac{u^2}{9.81} = 0.2.
Step 6: Rearranging this to find u, we multiply both sides by 9.81: u^2 = 0.2 \times 9.81.
Step 7: Thus, u^2 = 1.962, so u = \sqrt{1.962} \approx 1.4 \text{ m/s}.
Therefore, the minimum take-off speed of the grasshopper is approximately 1.4 m/s.