A stone is projected from the ground with a velocity of `14 ms ^(-1).` One second later it clears a wall 2m high. The angle of projection is `(g = 10 ms ^(-2))`

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `15^(@)`

Answer

To solve the problem, we need to find the angle of projection (θ) of a stone that is projected from the ground with a velocity of 14 m/s and clears a wall of height 2 m after 1 second. We can use the equations of motion for projectile motion to derive the angle.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity (U) = 14 m/s
- Height of the wall (h) = 2 m
- Time taken to clear the wall (t) = 1 s
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)

2. **Break down the initial velocity into components:**
- The vertical component of the initial velocity (U_y) is given by:
\[ U_y = U \sin \theta = 14 \sin \theta \]
- The horizontal component (U_x) is not needed for this calculation, but it is:
\[ U_x = U \cos \theta = 14 \cos \theta \]

3. **Use the vertical motion equation to find the height:**
- The vertical displacement (s) after time t is given by the equation:
\[ s = U_y t - \frac{1}{2} g t^2 \]
- Substituting the known values:
\[ 2 = (14 \sin \theta)(1) - \frac{1}{2} (10)(1^2) \]
- Simplifying this gives:
\[ 2 = 14 \sin \theta - 5 \]
- Rearranging the equation:
\[ 14 \sin \theta = 2 + 5 = 7 \]
- Therefore:
\[ \sin \theta = \frac{7}{14} = \frac{1}{2} \]

4. **Find the angle of projection:**
- The angle θ can be found using the inverse sine function:
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \]
- From trigonometric values, we know:
\[ \theta = 30^\circ \]

### Final Answer:
The angle of projection (θ) is **30 degrees**.