A ball is dropped from the top of a building of height 80 m. At same instant another ball is thrown upwards with speed 50 m/s from the bottom of the building. The time at which balls will meet is

Options

A. 1.6 s
B. 5 s
C. 8 s
D. 10 s

Answer

To solve the problem, we need to determine the time at which the two balls meet. We will use the equations of motion to find the distances traveled by each ball and set them equal to the total height of the building.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Height of the building (H) = 80 m
- Initial velocity of the ball dropped from the top (v1) = 0 m/s (since it is dropped)
- Initial velocity of the ball thrown upwards (v2) = 50 m/s
- Acceleration due to gravity (g) = 9.8 m/s² (approximately 10 m/s² for simplicity)

2. **Set Up the Equations of Motion:**
- For the ball dropped from the top:
- Distance fallen (h1) after time t:
\[ h_1 = v_1 t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
- For the ball thrown upwards:
- Distance traveled upwards (h2) after time t:
\[ h_2 = v_2 t - \frac{1}{2} g t^2 = 50t - \frac{1}{2} g t^2 \]

3. **Total Distance:**
- The sum of the distances traveled by both balls must equal the height of the building:
\[ h_1 + h_2 = H \]
- Substituting the expressions for h1 and h2:
\[ \frac{1}{2} g t^2 + (50t - \frac{1}{2} g t^2) = 80 \]

4. **Simplify the Equation:**
- The \(\frac{1}{2} g t^2\) terms cancel out:
\[ 50t = 80 \]

5. **Solve for Time (t):**
- Rearranging gives:
\[ t = \frac{80}{50} = 1.6 \text{ seconds} \]

### Final Answer:
The time at which the balls will meet is **1.6 seconds**.