A body falls freely from a height of 50 m. Simultaneously, another body is thrown from the surface of earth with a certain initial velocity. The two bodies meet at a height of 10 m. What is the initial velocity of the second body ?

Options

A. `4.9 m//s`
B. `9.8m//s`
C. `17.5m//s`
D. `19.6m//s `

Answer

To solve the problem, we need to analyze the motion of both bodies and set up equations based on their positions over time.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the initial velocity of the second body (thrown upwards) be \( v_0 \).
- The height from which the first body is dropped is \( h = 50 \, \text{m} \).
- The height at which both bodies meet is \( h_1 = 10 \, \text{m} \).
- The distance fallen by the first body when they meet is \( h_2 = 50 - 10 = 40 \, \text{m} \).
- The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).

2. **Equation of Motion for the Falling Body:**
- The first body is in free fall, so we can use the equation of motion:
\[ h_2 = \frac{1}{2} g t^2 \]
- Substituting \( h_2 = 40 \, \text{m} \):
\[ 40 = \frac{1}{2} \cdot 10 \cdot t^2 \]
- Simplifying this gives:
\[ 40 = 5 t^2 \implies t^2 = 8 \implies t = \sqrt{8} = 2\sqrt{2} \, \text{s} \]

3. **Equation of Motion for the Thrown Body:**
- The second body is thrown upwards with an initial velocity \( v_0 \). The height it reaches after time \( t \) is given by:
\[ h_1 = v_0 t - \frac{1}{2} g t^2 \]
- Substituting \( h_1 = 10 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 10 = v_0 (2\sqrt{2}) - \frac{1}{2} \cdot 10 \cdot (2\sqrt{2})^2 \]
- Simplifying the second term:
\[ 10 = v_0 (2\sqrt{2}) - \frac{1}{2} \cdot 10 \cdot 8 \]
\[ 10 = v_0 (2\sqrt{2}) - 40 \]
- Rearranging gives:
\[ v_0 (2\sqrt{2}) = 50 \implies v_0 = \frac{50}{2\sqrt{2}} = \frac{25}{\sqrt{2}} = 25\sqrt{2} \, \text{m/s} \]

4. **Final Calculation:**
- To get a numerical value:
\[ v_0 \approx 25 \times 1.414 \approx 35.35 \, \text{m/s} \]

### Conclusion:
The initial velocity of the second body is approximately \( 35.35 \, \text{m/s} \).