A person sitting in the ground floor of a building notices through window of height 1.5 m a ball dropped from roof of building crosses window in 0.1s what is velocity of ball when it is at the tomost point of window

Options

A. `15.5 m/s`
B. `14.5 m/s`
C. `4.5 m/s`
D. `20 m/s`

Answer

To solve the problem, we need to find the velocity of the ball when it is at the topmost point of the window. We know the height of the window (1.5 m) and the time it takes for the ball to cross this window (0.1 s).

### Step-by-step Solution:

1. **Understanding the Problem**:
- The ball is dropped from the roof of the building and crosses a window of height 1.5 m in 0.1 seconds. We need to find its velocity at the topmost point of the window.

2. **Using the Equation of Motion**:
- We can use the second equation of motion for uniformly accelerated motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Where:
- \( s \) = displacement (1.5 m)
- \( u \) = initial velocity (which we need to find)
- \( a \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( t \) = time (0.1 s)

3. **Substituting the Known Values**:
- Plugging in the values into the equation:
\[ 1.5 = u \cdot 0.1 + \frac{1}{2} \cdot 10 \cdot (0.1)^2 \]
- Simplifying the equation:
\[ 1.5 = 0.1u + \frac{1}{2} \cdot 10 \cdot 0.01 \]
\[ 1.5 = 0.1u + 0.05 \]

4. **Isolating the Initial Velocity \( u \)**:
- Rearranging the equation to solve for \( u \):
\[ 1.5 - 0.05 = 0.1u \]
\[ 1.45 = 0.1u \]
\[ u = \frac{1.45}{0.1} = 14.5 \, \text{m/s} \]

5. **Conclusion**:
- The velocity of the ball when it is at the topmost point of the window is \( 14.5 \, \text{m/s} \).