A ball is dropped from top of a tower of 100 m height. Simultaneously another ball was thrown upward from bottom of the tower with a speed of `50 m//s (g = 10m//s^(2))`. They will cross each other after

Options

A. 1s
B. 2s
C. 3s
D. 4s

Answer

To solve the problem of when the two balls will cross each other, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A ball is dropped from a height of 100 m (let's call this Ball A).
- Another ball is thrown upwards from the ground with an initial speed of 50 m/s (let's call this Ball B).
- We need to find the time `t` at which both balls meet.

2. **Setting Up the Equations**:
- For Ball A (dropped from the top):
- Initial velocity, \( u_A = 0 \) m/s (since it is dropped).
- Height fallen after time \( t \) is given by the equation:
\[ H_A = u_A t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} (10) t^2 = 5t^2 \]
- The height of Ball A above the ground after time \( t \) is:
\[ H_A = 100 - 5t^2 \]

- For Ball B (thrown upwards):
- Initial velocity, \( u_B = 50 \) m/s.
- Height reached after time \( t \) is given by the equation:
\[ H_B = u_B t - \frac{1}{2} g t^2 = 50t - \frac{1}{2} (10) t^2 = 50t - 5t^2 \]

3. **Setting the Heights Equal**:
- At the time of crossing, the heights of both balls will be equal:
\[ H_A = H_B \]
- Substituting the equations we derived:
\[ 100 - 5t^2 = 50t - 5t^2 \]

4. **Simplifying the Equation**:
- We can cancel \( -5t^2 \) from both sides:
\[ 100 = 50t \]

5. **Solving for Time \( t \)**:
- Rearranging gives:
\[ t = \frac{100}{50} = 2 \text{ seconds} \]

### Final Answer:
The two balls will cross each other after **2 seconds**.