A stone is launched upward at `45^(@)` with speed `v_(0)`. A bee follow the trajectory of the stone at a constant speed equal to the initial speed of the stone.
(a) Find the radius of curvature at the top point of the trajectory.
(b) What is the acceleration of the bee at the top point of the trajectory? For the stone, neglect the air resistance.

Answer

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Find the radius of curvature at the top point of the trajectory.

1. **Understanding the Motion**:
- The stone is launched at an angle of \(45^\circ\) with an initial speed \(v_0\).
- At the top point of the trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains.

2. **Calculating the Components of Velocity**:
- The horizontal component of the velocity \(v_x\) at the top point is given by:
\[ v_x = v_0 \cos(45^\circ) = \frac{v_0}{\sqrt{2}} \]
- The vertical component of the velocity \(v_y\) at the top point is:
\[ v_y = v_0 \sin(45^\circ) = \frac{v_0}{\sqrt{2}} \]
- At the top point, \(v_y = 0\) and \(v_x = \frac{v_0}{\sqrt{2}}\).

3. **Normal Acceleration at the Top Point**:
- The only acceleration acting on the stone at the top point is the gravitational acceleration \(g\) acting downward.
- The normal acceleration \(a_n\) is given by the formula:
\[ a_n = \frac{v^2}{R} \]
- Here, \(v\) is the horizontal velocity \(v_x\) at the top point.

4. **Setting Up the Equation**:
- We can equate the normal acceleration to gravitational acceleration:
\[ g = \frac{v_x^2}{R} \]
- Substituting \(v_x = \frac{v_0}{\sqrt{2}}\):
\[ g = \frac{\left(\frac{v_0}{\sqrt{2}}\right)^2}{R} \]
- Simplifying gives:
\[ g = \frac{v_0^2}{2R} \]

5. **Solving for Radius of Curvature \(R\)**:
- Rearranging the equation to solve for \(R\):
\[ R = \frac{v_0^2}{2g} \]

### Part (b): What is the acceleration of the bee at the top point of the trajectory?

1. **Understanding the Bee's Motion**:
- The bee follows the stone at a constant speed \(v_0\). Since it follows the trajectory, it experiences circular motion.

2. **Acceleration of the Bee**:
- The bee's acceleration at the top point is the normal acceleration because it is moving with constant speed along the curved path.
- The normal acceleration \(a_n\) is given by:
\[ a_n = \frac{v^2}{R} \]
- Substituting \(v = v_0\) and \(R = \frac{v_0^2}{2g}\):
\[ a_n = \frac{v_0^2}{\frac{v_0^2}{2g}} = 2g \]

### Final Answers:
- **(a)** The radius of curvature at the top point of the trajectory is:
\[ R = \frac{v_0^2}{2g} \]
- **(b)** The acceleration of the bee at the top point of the trajectory is:
\[ a = 2g \]