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)` from 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 will meet, we can break it down step by step.

### Step 1: Understand the problem
We have two balls:
- Ball 1 is dropped from the top of a 100 m high building (initial velocity \( u_1 = 0 \)).
- Ball 2 is thrown upwards from the bottom of the building with an initial velocity of \( u_2 = 40 \, \text{m/s} \).

### Step 2: Define the motion equations
For Ball 1 (dropped from the top):
- The distance fallen after time \( t \) can be given by the equation:
\[ h_1 = \frac{1}{2} g t^2 \]
where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

For Ball 2 (thrown upwards):
- The distance traveled upwards after time \( t \) can be given by the equation:
\[ h_2 = u_2 t - \frac{1}{2} g t^2 \]
where \( u_2 = 40 \, \text{m/s} \).

### Step 3: Set up the equation for the total height
The total height of the building is 100 m. Therefore, the sum of the distances covered by both balls when they meet will equal the height of the building:
\[ h_1 + h_2 = 100 \]
Substituting the equations we have:
\[ \frac{1}{2} g t^2 + (u_2 t - \frac{1}{2} g t^2) = 100 \]
This simplifies to:
\[ u_2 t = 100 \]

### Step 4: Substitute the values
Substituting \( u_2 = 40 \, \text{m/s} \):
\[ 40t = 100 \]

### Step 5: Solve for time \( t \)
Now, solving for \( t \):
\[ t = \frac{100}{40} = 2.5 \, \text{s} \]

### Conclusion
The two balls will meet after **2.5 seconds**.

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