A stone is projected from the ground with velocity `25m//s` . Two seconds later, it just clears a wall 5 m high. The angle of projection of the stone is `(g = 10m //sec^(2))`

Options

A. `30^(@)`
B. `45^(@)`
C. `50.2^(@)`
D. `60^(@)`

Answer

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the Problem
We have a stone projected from the ground with an initial velocity of \( 25 \, \text{m/s} \). After \( 2 \, \text{s} \), it clears a wall that is \( 5 \, \text{m} \) high. We need to find the angle of projection \( \theta \).

### Step 2: Identify Components of Motion
The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component: \( V_{x} = 25 \cos \theta \)
- Vertical component: \( V_{y} = 25 \sin \theta \)

### Step 3: Use the Vertical Motion Equation
We will use the vertical motion equation to find the height reached by the stone after \( 2 \, \text{s} \):
\[ S_y = V_{y} t - \frac{1}{2} g t^2 \]
Where:
- \( S_y = 5 \, \text{m} \) (height of the wall)
- \( V_{y} = 25 \sin \theta \)
- \( t = 2 \, \text{s} \)
- \( g = 10 \, \text{m/s}^2 \)

### Step 4: Substitute Known Values into the Equation
Substituting the known values into the equation:
\[ 5 = (25 \sin \theta)(2) - \frac{1}{2}(10)(2^2) \]
Calculating the second term:
\[ \frac{1}{2}(10)(2^2) = \frac{1}{2}(10)(4) = 20 \]
Thus, the equation becomes:
\[ 5 = 50 \sin \theta - 20 \]

### Step 5: Rearranging the Equation
Rearranging the equation to isolate \( \sin \theta \):
\[ 5 + 20 = 50 \sin \theta \]
\[ 25 = 50 \sin \theta \]
\[ \sin \theta = \frac{25}{50} = \frac{1}{2} \]

### Step 6: Find the Angle of Projection
The angle \( \theta \) for which \( \sin \theta = \frac{1}{2} \) is:
\[ \theta = 30^\circ \]

### Final Answer
The angle of projection of the stone is \( 30^\circ \).

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