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 an object being thrown at a speed of 20 m/s at an angle of 45 degrees above the horizontal, we need to find the time taken for the object to return to the same level from which it was thrown. We will use the equations of motion for projectile motion.

### Step-by-Step Solution:

1. **Identify the Initial Parameters:**
- Initial speed (u) = 20 m/s
- Angle of projection (θ) = 45 degrees
- Acceleration due to gravity (g) = 9.81 m/s² (approximately)

2. **Resolve the Initial Velocity:**
The initial velocity can be resolved into horizontal (u_x) and vertical (u_y) components:
\[ u_x = u \cos(θ) = 20 \cos(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]
\[ u_y = u \sin(θ) = 20 \sin(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

3. **Use the Time of Flight Formula:**
The time of flight (T) for a projectile launched and landing at the same height is given by:
\[ T = \frac{2u_y}{g} \]
Substituting the value of \(u_y\):
\[ T = \frac{2 \times 10\sqrt{2}}{9.81} \]

4. **Calculate the Time of Flight:**
\[ T = \frac{20\sqrt{2}}{9.81} \]
Now, calculate \(20\sqrt{2}\):
\[ 20\sqrt{2} \approx 20 \times 1.414 \approx 28.28 \]
Therefore:
\[ T \approx \frac{28.28}{9.81} \approx 2.89 \, \text{seconds} \]

5. **Final Answer:**
The time taken by the object to return to the same level is approximately **2.89 seconds**.