A ball is dropped from the top of a building the ball takes 0.5 s to fall past the 3 m length of a window at certain distance from the top of the building. Speed of the ball as it crosses the top edge of the window is `(g=10 m//s^(2))`

Options

A. `3.5 ms^(-1)`
B. `8.5 ms^(-1)`
C. `5 ms^(-1)`
D. `12 ms^(-1)`

Explanation

To solve the problem step by step, we can follow the kinematic equations of motion. Let's break it down:

### Step 1: Understand the Problem
A ball is dropped from the top of a building and falls past a window of height 3 meters in 0.5 seconds. We need to find the speed of the ball as it crosses the top edge of the window.

### Step 2: Identify Given Data
- Height of the window (s) = 3 m
- Time taken to pass the window (t) = 0.5 s
- Acceleration due to gravity (g) = 10 m/s²
- Initial velocity at the top of the window (u) = v (unknown)

### Step 3: Use the Kinematic Equation
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the known values:
- \( s = 3 \, \text{m} \)
- \( u = v \)
- \( a = g = 10 \, \text{m/s}^2 \)
- \( t = 0.5 \, \text{s} \)

The equation becomes:
\[ 3 = v(0.5) + \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]

### Step 4: Simplify the Equation
Calculating the second term:
\[ \frac{1}{2} \cdot 10 \cdot (0.5)^2 = \frac{1}{2} \cdot 10 \cdot 0.25 = \frac{10}{8} = 1.25 \]

Now substitute this back into the equation:
\[ 3 = 0.5v + 1.25 \]

### Step 5: Solve for v
Rearranging the equation:
\[ 0.5v = 3 - 1.25 \]
\[ 0.5v = 1.75 \]

Now, multiply both sides by 2 to isolate v:
\[ v = 1.75 \times 2 \]
\[ v = 3.5 \, \text{m/s} \]

### Final Answer
The speed of the ball as it crosses the top edge of the window is **3.5 m/s**.

---