A small object is dropped from the top of a building and- falls to the ground. As it falls, accelerating due to gravity, it passes window. It has speed `v_(1)` at the top of the window and speed `v_(2)` at the bottom of the window, at what point does it have speed `(v_(1)+v_(2))/2` ? Neglect the air resistance.

Options

A. It depends on the height of the window or its distance from the top of the building.
B. Above the centre point of the window
C. Below the centre point of the window
D. At the centre point of the window

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have an object falling freely under gravity, passing through a window of height \( x \). It has a speed \( v_1 \) at the top of the window and a speed \( v_2 \) at the bottom of the window. We need to find the point at which its speed is \( \frac{v_1 + v_2}{2} \).

### Step 2: Use the kinematic equation
We can use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity
- \( u \) is the initial velocity
- \( a \) is the acceleration (which is \( g = 10 \, \text{m/s}^2 \))
- \( s \) is the displacement

### Step 3: Apply the equation for \( v_2 \)
For the object falling from the top of the window to the bottom, we can write:
\[ v_2^2 = v_1^2 + 2g x \]
This can be rearranged to:
\[ v_2^2 - v_1^2 = 20x \quad \text{(1)} \]

### Step 4: Find the point where speed is \( \frac{v_1 + v_2}{2} \)
Let’s denote the height from the top of the window to the point where the speed is \( \frac{v_1 + v_2}{2} \) as \( h \). We can apply the kinematic equation again:
\[ \left(\frac{v_1 + v_2}{2}\right)^2 = v_1^2 + 2g h \]
Expanding the left side gives:
\[ \frac{(v_1 + v_2)^2}{4} = v_1^2 + 20h \quad \text{(2)} \]

### Step 5: Expand and rearrange equation (2)
Expanding the left side:
\[ \frac{v_1^2 + 2v_1v_2 + v_2^2}{4} = v_1^2 + 20h \]
Multiplying through by 4 to eliminate the fraction:
\[ v_1^2 + 2v_1v_2 + v_2^2 = 4v_1^2 + 80h \]
Rearranging gives:
\[ 2v_1v_2 + v_2^2 - 3v_1^2 = 80h \quad \text{(3)} \]

### Step 6: Substitute equation (1) into equation (3)
From equation (1), we know:
\[ v_2^2 - v_1^2 = 20x \implies v_2^2 = v_1^2 + 20x \]
Substituting this into equation (3):
\[ 2v_1v_2 + (v_1^2 + 20x) - 3v_1^2 = 80h \]
This simplifies to:
\[ 2v_1v_2 - 2v_1^2 + 20x = 80h \]
Rearranging gives:
\[ 20x = 80h - 2v_1(v_1 - v_2) \]

### Step 7: Solve for \( h \)
Now, we can express \( h \):
\[ h = \frac{20x + 2v_1(v_1 - v_2)}{80} \]
This indicates that the point where the speed is \( \frac{v_1 + v_2}{2} \) is below the center of the window because \( h \) is less than \( \frac{x}{2} \).

### Conclusion
Thus, the object reaches the speed \( \frac{v_1 + v_2}{2} \) below the center point of the window.