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 step by step, we will first analyze the motion of the stone projected from the ground. We will break down the initial velocity into its horizontal and vertical components and then use the equations of motion to find the angle of projection.

### Step 1: Identify the given data
- Initial velocity (u) = 14 m/s
- Height of the wall (h) = 2 m
- Time taken to reach the wall (t) = 1 s
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Break down the initial velocity into components
The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component (u_x) = u * cos(θ)
- Vertical component (u_y) = u * sin(θ)

Where θ is the angle of projection.

### Step 3: Use the vertical motion equation
The vertical displacement (y) after time t can be expressed using the equation of motion:
\[ y = u_y \cdot t - \frac{1}{2} g t^2 \]

Substituting the known values:
\[ 2 = (14 \sin(θ)) \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 \]

### Step 4: Simplify the equation
This simplifies to:
\[ 2 = 14 \sin(θ) - 5 \]
\[ 14 \sin(θ) = 2 + 5 \]
\[ 14 \sin(θ) = 7 \]
\[ \sin(θ) = \frac{7}{14} = \frac{1}{2} \]

### Step 5: Determine the angle of projection
From the value of sin(θ):
\[ \sin(θ) = \frac{1}{2} \]
This implies:
\[ θ = 30^\circ \]

### Conclusion
The angle of projection is \( 30^\circ \).

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