A body thrown up with a finite speed is caught back after 4 sec. The speed of the body with which it is thrown up is

Options

A. 10 m/sec
B. 20 m/sec
C. 30 m/sec
D. 40 m/sec

Answer

To solve the problem of finding the speed with which a body is thrown upwards and caught back after 4 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a body is thrown upwards, it will rise to a certain height and then fall back down. The total time taken for this journey is given as 4 seconds.

2. **Identifying the Variables**:
- Let \( u \) be the initial speed with which the body is thrown upwards.
- The total time of flight \( t = 4 \) seconds.
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) (acting downwards).

3. **Using the Equation of Motion**:
- The displacement \( s \) when the body returns to the original position is 0.
- We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s = 0 \), \( a = -g = -10 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \).

4. **Substituting the Values**:
- Plugging in the values into the equation:
\[ 0 = u(4) + \frac{1}{2}(-10)(4^2) \]
- This simplifies to:
\[ 0 = 4u - 5 \cdot 16 \]

5. **Solving for \( u \)**:
- Rearranging the equation gives:
\[ 4u = 80 \]
- Dividing both sides by 4:
\[ u = 20 \, \text{m/s} \]

6. **Conclusion**:
- The speed of the body with which it is thrown upwards is \( 20 \, \text{m/s} \).