A rock is launched upward at `45^@`. A bee moves along the trajectory of the rock. What is the magnitude of acceleration `("in" m s^-2)` of the bee at the top point of the trajectory ? For the rock, neglect the air resistance.

Explanation

To solve the problem, we need to determine the magnitude of acceleration of the bee at the top point of the trajectory of the rock. Here’s a step-by-step solution:

### Step 1: Understand the motion of the rock
When the rock is launched at an angle of \(45^\circ\), it follows a parabolic trajectory. At the top of this trajectory, the vertical component of the velocity becomes zero, but the horizontal component remains.

### Step 2: Identify the forces acting on the bee
Since the bee is moving along the trajectory of the rock and we are neglecting air resistance, the only force acting on the bee at the top of the trajectory is the gravitational force. Therefore, the acceleration of the bee will be equal to the acceleration due to gravity.

### Step 3: Determine the acceleration due to gravity
The acceleration due to gravity \(g\) is approximately \(9.81 \, \text{m/s}^2\). This is the acceleration that acts downward on both the rock and the bee at the top of the trajectory.

### Step 4: Conclusion
Since the bee is moving along the trajectory of the rock and is affected only by gravity at the top point, the magnitude of the acceleration of the bee at the top point of the trajectory is:
\[ \text{Magnitude of acceleration of the bee} = g = 9.81 \, \text{m/s}^2 \]

### Final Answer
The magnitude of acceleration of the bee at the top point of the trajectory is \(9.81 \, \text{m/s}^2\).

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