A piece of marble is projected from the earth's surface with a velocity of `50 m s^(-1)` . 2 s later , it just clears a wall 5 m high. What is the angle of projection ?

Options

A. `45^@`
B. `30^@`
C. `60^@`
D. `sin ^(-1)(1/4)`

Answer

To solve the problem step by step, we need to analyze the motion of the marble projected from the Earth's surface.

### Step 1: Understand the problem
We have a marble projected with an initial velocity \( u = 50 \, \text{m/s} \). After \( t = 2 \, \text{s} \), it clears a wall of height \( h = 5 \, \text{m} \). We need to find the angle of projection \( \theta \).

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

### Step 3: Apply the equation of motion for vertical displacement
The vertical displacement \( S \) after time \( t \) can be calculated using the equation of motion:
\[ S = u_y t + \frac{1}{2} a t^2 \]
Here, \( a \) is the acceleration due to gravity, which acts downwards and is \( -g \) (taking \( g = 10 \, \text{m/s}^2 \)).

Substituting the values:
\[ 5 = (u \sin \theta)(2) - \frac{1}{2} (10)(2^2) \]
This simplifies to:
\[ 5 = 2u \sin \theta - 20 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2u \sin \theta = 25 \]
Substituting \( u = 50 \, \text{m/s} \):
\[ 2(50) \sin \theta = 25 \]
\[ 100 \sin \theta = 25 \]

### Step 5: Solve for \( \sin \theta \)
Dividing both sides by 100:
\[ \sin \theta = \frac{25}{100} = \frac{1}{4} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{1}{4}\right) \]

### Final Answer
Thus, the angle of projection \( \theta \) is:
\[ \theta = \sin^{-1}\left(\frac{1}{4}\right) \]