A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of `40ms^(-1)` form the bottom of the building. The two balls will meet after.

Options

A. 5 s
B. 2.5 s
C. 2 s
D. 3 s

Explanation

To solve the problem of when the two balls meet, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- A ball is dropped from the top of a 100 m high building (let's call this Ball A).
- Another ball is thrown upwards from the ground with an initial velocity of 40 m/s (let's call this Ball B).
- We need to find the time \( t \) when both balls meet.

2. **Define the Variables**:
- Let \( t \) be the time in seconds after the balls are released.
- The height of the building is \( h = 100 \) m.
- The acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \).

3. **Calculate the Distance Traveled by Ball A**:
- Ball A is dropped from rest, so its initial velocity \( u_A = 0 \).
- The distance \( S_1 \) traveled by Ball A after time \( t \) can be calculated using the equation:
\[ S_1 = u_A t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g t^2 = \frac{1}{2} (9.81) t^2 = 4.905 t^2 \]

4. **Calculate the Distance Traveled by Ball B**:
- Ball B is thrown upwards with an initial velocity of \( u_B = 40 \, \text{m/s} \).
- The distance \( S_2 \) traveled by Ball B after time \( t \) can be calculated using the equation:
\[ S_2 = u_B t - \frac{1}{2} g t^2 = 40t - \frac{1}{2} (9.81) t^2 = 40t - 4.905 t^2 \]

5. **Set Up the Equation**:
- Since the total distance between the two balls must equal the height of the building, we can write:
\[ S_1 + S_2 = 100 \]
- Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ 4.905 t^2 + (40t - 4.905 t^2) = 100 \]

6. **Simplify the Equation**:
- The \( 4.905 t^2 \) terms cancel out:
\[ 40t = 100 \]

7. **Solve for \( t \)**:
- Dividing both sides by 40:
\[ t = \frac{100}{40} = 2.5 \, \text{s} \]

### Final Answer:
The two balls will meet after **2.5 seconds**.