A body throws a ball from shoulder height at an initial velocity of 30 m/s. Spending 4.8 s in air, the ball is caught by another boy as the same shoulder-height level. What is the angle of projection ?

Options

A. `37^(@)`
B. `30^(@)`
C. `53^(@)`
D. `60^(@)`

Answer

To find the angle of projection of the ball thrown from shoulder height, we can use the following steps:

### Step-by-Step Solution:

1. **Understand the Problem:**
The ball is thrown with an initial velocity \( u = 30 \, \text{m/s} \) and remains in the air for a total time \( T = 4.8 \, \text{s} \). The ball is caught at the same height from which it was thrown.

2. **Use the Time of Flight Formula:**
The time of flight for a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

3. **Substitute Known Values:**
Plugging in the known values into the time of flight formula:
\[ 4.8 = \frac{2 \times 30 \times \sin \theta}{10} \]

4. **Simplify the Equation:**
Rearranging the equation gives:
\[ 4.8 = \frac{60 \sin \theta}{10} \]
\[ 4.8 = 6 \sin \theta \]

5. **Solve for \( \sin \theta \):**
Dividing both sides by 6:
\[ \sin \theta = \frac{4.8}{6} = 0.8 \]

6. **Find the Angle \( \theta \):**
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}(0.8) \]
Using a calculator, we find:
\[ \theta \approx 53^\circ \]

7. **Conclusion:**
The angle of projection is approximately \( 53^\circ \).

### Final Answer:
The angle of projection is \( 53^\circ \).

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