A particle projected at some angle with velocity 50 m/s crosses a 20 m high wall after 4 s from the time of projection. The angle of porjection of the particle is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `53^(@)`

Answer

To solve the problem step by step, we will follow the projectile motion equations and the given conditions.

### Step 1: Identify the components of the initial velocity
The initial velocity \( u \) of the particle is given as 50 m/s. This velocity can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta = 50 \cos \theta \)
- Vertical component: \( u_y = u \sin \theta = 50 \sin \theta \)

### Step 2: Use the vertical motion equation
The vertical motion of the particle can be described using the equation:
\[ S = ut + \frac{1}{2} a t^2 \]
where:
- \( S \) is the vertical displacement (height of the wall = 20 m),
- \( u \) is the initial vertical velocity,
- \( t \) is the time of flight (4 s),
- \( a \) is the acceleration due to gravity (approximately -10 m/s², acting downwards).

Substituting the known values into the equation:
\[ 20 = (50 \sin \theta)(4) + \frac{1}{2}(-10)(4^2) \]

### Step 3: Simplify the equation
Calculating the second term:
\[ \frac{1}{2}(-10)(16) = -80 \]
Now substituting this back into the equation:
\[ 20 = 200 \sin \theta - 80 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 200 \sin \theta = 20 + 80 \]
\[ 200 \sin \theta = 100 \]

### Step 5: Solve for \( \sin \theta \)
Now, divide both sides by 200:
\[ \sin \theta = \frac{100}{200} = \frac{1}{2} \]

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

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