An object is being thrown at a speed of 20 m/s in a direction `45^(@)` above the horizontal . The time taken by the object to return to the same level is

Options

A. 20 /g
B. 20 g
C. `20 sqrt(2)//g`
D. `20sqrt(2)`g

Answer

To solve the problem of finding the time taken by an object thrown at a speed of 20 m/s at an angle of 45 degrees above the horizontal to return to the same level, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial speed, \( u = 20 \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) (approximately)

2. **Use the Formula for Time of Flight:**
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

3. **Calculate \( \sin \theta \):**
Since \( \theta = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]

4. **Substitute the Values into the Formula:**
Plugging in the values we have:
\[ T = \frac{2 \times 20 \times \sin 45^\circ}{g} \]
\[ T = \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{9.81} \]

5. **Simplify the Expression:**
\[ T = \frac{40 \times \frac{1}{\sqrt{2}}}{9.81} \]
\[ T = \frac{40}{9.81 \sqrt{2}} \]

6. **Calculate the Numerical Value:**
First, calculate \( \sqrt{2} \approx 1.414 \):
\[ T \approx \frac{40}{9.81 \times 1.414} \approx \frac{40}{13.888} \approx 2.88 \, \text{s} \]

7. **Final Result:**
The time taken by the object to return to the same level is approximately \( 2.88 \, \text{s} \).