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, given that it crosses a window of height 1.5 m in 0.1 seconds.

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Height of the window (\( \Delta s \)) = 1.5 m
- Time taken to cross the window (\( t \)) = 0.1 s
- Acceleration due to gravity (\( g \)) = 10 m/s² (approximated)

2. **Use the Equation of Motion**:
We can use the equation of motion:
\[ \Delta s = ut + \frac{1}{2} a t^2 \]
where:
- \( \Delta s \) = height of the window (1.5 m)
- \( u \) = initial velocity of the ball when it enters the window
- \( a \) = acceleration (which is \( g = 10 \, \text{m/s}^2 \))
- \( t \) = time (0.1 s)

3. **Substitute the Known Values**:
Substitute the known values into the equation:
\[ 1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2 \]

4. **Calculate the Second Term**:
Calculate \( \frac{1}{2}(10)(0.1)^2 \):
\[ \frac{1}{2}(10)(0.01) = 0.05 \]
So the equation becomes:
\[ 1.5 = 0.1u + 0.05 \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ 0.1u = 1.5 - 0.05 \]
\[ 0.1u = 1.45 \]

6. **Solve for \( u \)**:
Divide both sides by 0.1:
\[ u = \frac{1.45}{0.1} = 14.5 \, \text{m/s} \]

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