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

Explanation

To solve the problem step by step, we will use the equations of motion. Here’s how we can find the speed of the ball as it crosses the top edge of the window.

### Step 1: Understand the given information
- The ball is dropped from the top of a building.
- The time taken to fall past the 3 m length of the window is \( t = 0.5 \) seconds.
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
- The initial velocity \( u = 0 \) (since the ball is dropped).

### Step 2: Use the equation of motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
Where:
- \( s \) is the displacement (3 m, the length of the window),
- \( u \) is the initial velocity (0 m/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( t \) is the time (0.5 s).

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ 3 = 0 \cdot 0.5 + \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]
This simplifies to:
\[ 3 = \frac{1}{2} \cdot 10 \cdot 0.25 \]
\[ 3 = \frac{10}{8} = 1.25 \]
This indicates that the displacement of the ball while passing the window is indeed 3 m.

### Step 4: Calculate the final velocity at the top edge of the window
To find the speed of the ball as it crosses the top edge of the window, we can use the first equation of motion:
\[ v = u + gt \]
Where:
- \( v \) is the final velocity,
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \),
- \( t = 0.5 \, \text{s} \).

Substituting the values:
\[ v = 0 + 10 \cdot 0.5 \]
\[ v = 5 \, \text{m/s} \]

### Step 5: Calculate the speed at the bottom edge of the window
Now we need to find the speed at the bottom edge of the window. For this, we can use the same equation of motion:
\[ v^2 = u^2 + 2gs \]
Where \( s = 3 \, \text{m} \) (the distance of the window). The initial velocity \( u \) at the top edge of the window is \( 5 \, \text{m/s} \).

Substituting the values:
\[ v^2 = 5^2 + 2 \cdot 10 \cdot 3 \]
\[ v^2 = 25 + 60 \]
\[ v^2 = 85 \]
\[ v = \sqrt{85} \approx 9.22 \, \text{m/s} \]

### Final Answer
The speed of the ball as it crosses the top edge of the window is approximately \( 9.22 \, \text{m/s} \).

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