The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is

Options

A. g
B. `(g)/(2)`
C. `(g)/(sqrt(2))`
D. `sqrt(2)g`

Answer

To solve the problem step-by-step, we need to analyze the information given about the freely falling body and use the equations of motion.

### Step 1: Understand the given information
We know that the average velocity \( V_{avg} \) of a freely falling body is numerically equal to half of the acceleration due to gravity \( g \). Therefore, we can write:
\[ V_{avg} = \frac{g}{2} \]

### Step 2: Relate average velocity to displacement and time
The average velocity can also be defined as the total displacement divided by the total time taken. If the body falls from a height \( h \) in time \( t \), we can express this as:
\[ V_{avg} = \frac{h}{t} \]

### Step 3: Set up the equation
From the information given, we can equate the two expressions for average velocity:
\[ \frac{h}{t} = \frac{g}{2} \]
From this, we can express the height \( h \) in terms of \( t \):
\[ h = \frac{g}{2} t \]

### Step 4: Use the equation of motion
For a freely falling body starting from rest, we can use the second equation of motion:
\[ h = ut + \frac{1}{2} g t^2 \]
Since the initial velocity \( u = 0 \), this simplifies to:
\[ h = \frac{1}{2} g t^2 \]

### Step 5: Equate the two expressions for height
Now we have two expressions for \( h \):
1. \( h = \frac{g}{2} t \)
2. \( h = \frac{1}{2} g t^2 \)

Setting these equal to each other gives:
\[ \frac{g}{2} t = \frac{1}{2} g t^2 \]

### Step 6: Simplify the equation
We can cancel \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ t = t^2 \]
This implies:
\[ t^2 - t = 0 \]
Factoring gives:
\[ t(t - 1) = 0 \]
Thus, \( t = 0 \) or \( t = 1 \). Since \( t = 0 \) is not a valid solution in this context, we have:
\[ t = 1 \text{ second} \]

### Step 7: Find the final velocity
Now, we can find the final velocity \( v \) of the body as it reaches the ground using the first equation of motion:
\[ v = u + gt \]
Substituting \( u = 0 \) and \( t = 1 \):
\[ v = 0 + g \cdot 1 = g \]

### Conclusion
The velocity of the body as it reaches the ground is:
\[ \boxed{g} \]