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

Answer

To solve the problem of when the two balls meet, we can break it down into a few steps.

### Step 1: Define the variables
- Let the height of the building be \( H = 100 \, \text{m} \).
- Let the time taken for the balls to meet be \( t \).
- The first ball (dropped from the top) has an initial velocity \( u_1 = 0 \, \text{m/s} \) and falls under gravity \( g = 9.8 \, \text{m/s}^2 \).
- The second ball (thrown upwards) has an initial velocity \( u_2 = 40 \, \text{m/s} \).

### Step 2: Write the equations of motion
For the first ball (dropped from the top):
- The distance fallen by the first ball after time \( t \) is given by:
\[ h_1 = u_1 t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
Thus, the distance fallen by the first ball is:
\[ h_1 = \frac{1}{2} g t^2 \]

For the second ball (thrown upwards):
- The distance covered by the second ball after time \( t \) is given by:
\[ h_2 = u_2 t - \frac{1}{2} g t^2 = 40t - \frac{1}{2} g t^2 \]
Since the total height of the building is 100 m, the distance from the bottom to the meeting point is:
\[ h_2 = 100 - h_1 \]

### Step 3: Set up the equation
Since both balls meet at the same height, we can equate the two distances:
\[ h_1 + h_2 = 100 \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{1}{2} g t^2 + (40t - \frac{1}{2} g t^2) = 100 \]
This simplifies to:
\[ 40t = 100 \]

### Step 4: Solve for \( t \)
Now, solve for \( t \):
\[ t = \frac{100}{40} = 2.5 \, \text{s} \]

### Conclusion
The two balls will meet after \( 2.5 \) seconds.

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