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Sagot :
Answer:To determine the minimum initial speed \( v_0 \) required for the cricket to jump over a pipe of diameter \( D \), we can analyze the projectile motion of the cricket's jump.
### Assumptions:
1. The cricket launches itself with an initial angle \( \theta \) above the horizontal.
2. The highest point of its trajectory should be at least at the level of the top of the pipe for it to clear the pipe.
### Setup:
- Let \( h \) be the vertical distance from the ground to the top of the pipe.
- The pipe has a diameter \( D \), so its radius \( R \) is \( \frac{D}{2} \).
### Conditions for clearing the pipe:
1. **Vertical Motion:**
At the highest point of the jump, the vertical position \( y_{\text{max}} \) should satisfy:
\[ y_{\text{max}} \geq h + R \]
Since \( y_{\text{max}} = h + v_0^2 \sin^2 \theta / (2g) \), where \( g \) is the acceleration due to gravity, we get:
\[ h + \frac{v_0^2 \sin^2 \theta}{2g} \geq h + \frac{D}{2} \]
2. **Horizontal Motion:**
The horizontal distance \( x \) the cricket travels before landing should be at least \( D \):
\[ x = \frac{v_0^2 \sin 2\theta}{g} \geq D \]
### Minimum speed calculation:
From the horizontal motion condition:
\[ v_0^2 \sin 2\theta \geq Dg \]
To find the minimum \( v_0 \), we can choose \( \theta \) that satisfies both conditions. However, for the minimum initial speed \( v_0 \), we consider the limiting case where \( \theta \) is such that the projectile just clears the pipe at the highest point of its trajectory.
Therefore, the minimum initial speed \( v_0 \) required for the cricket to clear the pipe of diameter \( D \) is given by:
\[ v_0 = \sqrt{\frac{Dg}{\sin 2\theta}} \]
where \( \theta \) is the launch angle that satisfies both the vertical and horizontal clearance conditions.
Explanation:
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