A body is vertically projected at 100 `ms^(-1)`. It returns after (`g=10 ms^(-2)`)

Options

A. 10s
B. 20s
C. 8s
D. 16s

Explanation

To solve the problem of a body projected vertically upwards with an initial velocity of 100 m/s and returning to its original position under the influence of gravity (g = 10 m/s²), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**: The body is projected upwards with an initial velocity (u) of 100 m/s. It will rise to a maximum height and then fall back down to the original position.

2. **Use the Equation of Motion**: We can use the second equation of motion, which is:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) is the displacement (which will be 0 when the body returns to the starting point),
- \(u\) is the initial velocity (100 m/s),
- \(a\) is the acceleration (which will be -g = -10 m/s², since gravity acts downwards),
- \(t\) is the time taken.

3. **Set Up the Equation**: Since the total displacement (s) when the body returns to the original position is 0, we can write:
\[ 0 = ut + \frac{1}{2} (-g) t^2 \]
Substituting the values:
\[ 0 = 100t - \frac{1}{2} (10) t^2 \]

4. **Simplify the Equation**: This simplifies to:
\[ 0 = 100t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 100t = 0 \]

5. **Factor the Equation**: Factoring out \(t\):
\[ t(5t - 100) = 0 \]
This gives us two solutions:
- \(t = 0\) (the time at which the body is projected)
- \(5t - 100 = 0\) which simplifies to \(t = 20\) seconds.

6. **Conclusion**: The time taken for the body to return to the original position is 20 seconds.

### Final Answer:
The body returns after **20 seconds**.