A body is projected horizontally from a very high tower with speed `20 ms ^(-1)` The approximate displacement of the body after 5 s is :

Options

A. 80 m
B. 120 m
C. 160 m
D. 320 m

Explanation

To solve the problem of a body projected horizontally from a tower, we will break it down into steps to find the displacement after 5 seconds.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The body is projected horizontally with an initial speed \( v_x = 20 \, \text{m/s} \).
- The time of flight \( t = 5 \, \text{s} \).

2. **Calculate Horizontal Displacement**:
- Since there is no acceleration in the horizontal direction, the horizontal displacement \( x \) can be calculated using the formula:
\[ x = v_x \cdot t \]
- Substituting the values:
\[ x = 20 \, \text{m/s} \times 5 \, \text{s} = 100 \, \text{m} \]

3. **Calculate Vertical Displacement**:
- The vertical displacement \( y \) can be calculated using the equation of motion:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
- Here, \( u_y = 0 \) (initial vertical velocity) and \( a_y = g \) (acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)).
- Thus, the equation simplifies to:
\[ y = \frac{1}{2} g t^2 \]
- Substituting the values:
\[ y = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (5 \, \text{s})^2 = \frac{1}{2} \times 9.81 \times 25 = 122.625 \, \text{m} \]
- We can round this to \( y \approx 125 \, \text{m} \) for simplicity.

4. **Calculate the Resultant Displacement**:
- The resultant displacement \( s \) can be found using the Pythagorean theorem:
\[ s = \sqrt{x^2 + y^2} \]
- Substituting the values of \( x \) and \( y \):
\[ s = \sqrt{(100 \, \text{m})^2 + (125 \, \text{m})^2} = \sqrt{10000 + 15625} = \sqrt{25625} \]
- Calculating this gives:
\[ s \approx 160 \, \text{m} \]

5. **Conclusion**:
- The approximate displacement of the body after 5 seconds is \( 160 \, \text{m} \).