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 finding the time taken by a cricket ball thrown at a speed of 30 m/s at an angle of 30 degrees above the horizontal to return to the same level, we can follow these steps:

### Step 1: Identify the given values
- Initial speed (u) = 30 m/s
- Angle of projection (θ) = 30 degrees
- Acceleration due to gravity (g) = 10 m/s² (approximately)

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

### Step 3: Calculate the sine of the angle
We need to calculate \(\sin 30^\circ\):
\[ \sin 30^\circ = \frac{1}{2} \]

### Step 4: Substitute the values into the formula
Now, substituting the values into the time of flight formula:
\[ T = \frac{2 \times 30 \times \sin 30^\circ}{10} \]
\[ T = \frac{2 \times 30 \times \frac{1}{2}}{10} \]

### Step 5: Simplify the expression
\[ T = \frac{30}{10} \]
\[ T = 3 \text{ seconds} \]

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

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