Find the average velocity of a particle released from rest from a height of `125 m` over a time interval till it strikes the ground. Take `g=10 m//s^2.`

Explanation

To find the average velocity of a particle released from rest from a height of 125 m, we can follow these steps:

### Step 1: Identify the known values
- Initial height (h) = 125 m
- Initial velocity (u) = 0 m/s (since the particle is released from rest)
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)

### Step 2: Determine the displacement
Since the particle is released from a height of 125 m and falls to the ground, the total displacement (s) is:
\[ s = -125 \, \text{m} \]
(The negative sign indicates that the displacement is downward.)

### Step 3: Use the kinematic equation to find the time taken (t)
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ -125 = 0 \cdot t + \frac{1}{2} (-10) t^2 \]
This simplifies to:
\[ -125 = -5 t^2 \]
\[ t^2 = \frac{125}{5} \]
\[ t^2 = 25 \]
Taking the square root of both sides gives:
\[ t = 5 \, \text{s} \]
(Since time cannot be negative, we take the positive root.)

### Step 4: Calculate the average velocity
The average velocity (v_avg) is given by the formula:
\[ v_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
Substituting the values we have:
\[ v_{\text{avg}} = \frac{-125 \, \text{m}}{5 \, \text{s}} \]
\[ v_{\text{avg}} = -25 \, \text{m/s} \]

### Step 5: Interpret the result
The negative sign indicates that the average velocity is directed downward.

### Final Answer
The average velocity of the particle is:
\[ v_{\text{avg}} = -25 \, \text{m/s} \]

---