A body is projected horizontally from the top of a hill with a velocity of 9.8 m/s. What time elapses before the vertical velocity is twice the horizontal velocity?

Options

A. 0.5 sec
B. 1 sec
C. 2 se
D. 1.5 sec

Answer

To solve the problem, we need to determine the time it takes for the vertical velocity of a body projected horizontally to become twice the horizontal velocity.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Horizontal velocity, \( V_{horizontal} = 9.8 \, \text{m/s} \)
- The condition we need to satisfy: \( V_{vertical} = 2 \times V_{horizontal} = 2 \times 9.8 \, \text{m/s} = 19.6 \, \text{m/s} \)

2. **Understand the Motion**:
- The body is projected horizontally, so its initial vertical velocity \( V_{vertical, initial} = 0 \, \text{m/s} \).
- The only force acting on the body in the vertical direction is gravity, which provides a constant acceleration \( g = 9.8 \, \text{m/s}^2 \) downward.

3. **Use the Kinematic Equation**:
- We can use the kinematic equation for vertical motion:
\[ V_{vertical} = V_{vertical, initial} + a \cdot t \]
- Here, \( V_{vertical} = 19.6 \, \text{m/s} \), \( V_{vertical, initial} = 0 \, \text{m/s} \), and \( a = g = 9.8 \, \text{m/s}^2 \).

4. **Substituting the Values**:
- Substitute the known values into the equation:
\[ 19.6 = 0 + 9.8 \cdot t \]

5. **Solve for Time \( t \)**:
- Rearranging the equation gives:
\[ t = \frac{19.6}{9.8} \]
- Calculate \( t \):
\[ t = 2 \, \text{seconds} \]

### Final Answer:
The time that elapses before the vertical velocity is twice the horizontal velocity is **2 seconds**.

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