A cricket ball is thrown at a speed of 30 m `s^(-1)` in a direction `30^(@)` above the horizontal. The time taken by the ball to return to the same level is

Options

A. 2 s
B. 3 s
C. 4 s
D. 5 s

Explanation

To solve the problem of a cricket ball thrown at a speed of 30 m/s at an angle of 30 degrees above the horizontal, we need to determine the time taken by the ball to return to the same level. This is known as the time of flight.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial speed (U) = 30 m/s
- Angle of projection (θ) = 30 degrees
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplicity)

2. **Use the Formula for Time of Flight:**
The formula for the time of flight (T) for a projectile launched from the ground level is:
\[ T = \frac{2U \sin \theta}{g} \]

3. **Calculate the Sine of the Angle:**
We need to find \(\sin(30^\circ)\):
\[ \sin(30^\circ) = \frac{1}{2} \]

4. **Substitute the Values into the Formula:**
Now, substituting the values into the time of flight formula:
\[ T = \frac{2 \times 30 \times \frac{1}{2}}{10} \]

5. **Simplify the Equation:**
Simplifying the equation:
\[ T = \frac{2 \times 30 \times 0.5}{10} = \frac{30}{10} = 3 \text{ seconds} \]

6. **Conclusion:**
The time taken by the ball to return to the same level is **3 seconds**.

### Final Answer:
The time taken by the ball to return to the same level is **3 seconds**.

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