A particle is projected from the earth's surface with a velocity of `"50 m s"^(-1)` at an angle `theta` with the horizontal. After `2 s` it just clears a wall `5 m` high. What is the value of `55 sin theta? (g = 10 ms^(-2))`

Answer

To solve the problem, we will use the equations of motion to find the value of \( 55 \sin \theta \).

### Step 1: Identify the given values
- Initial velocity \( u = 50 \, \text{m/s} \)
- Time \( t = 2 \, \text{s} \)
- Height of the wall \( S = 5 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

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

Thus, we have:
- \( u_y = 50 \sin \theta \)

### Step 3: Use the equation of motion for vertical displacement
The vertical displacement \( S \) can be expressed using the equation of motion:
\[ S = u_y t - \frac{1}{2} g t^2 \]
Substituting the known values:
\[ 5 = (50 \sin \theta)(2) - \frac{1}{2} (10)(2^2) \]

### Step 4: Simplify the equation
Substituting the values into the equation:
\[ 5 = 100 \sin \theta - \frac{1}{2} (10)(4) \]
\[ 5 = 100 \sin \theta - 20 \]

### Step 5: Solve for \( \sin \theta \)
Rearranging the equation gives:
\[ 100 \sin \theta = 5 + 20 \]
\[ 100 \sin \theta = 25 \]
\[ \sin \theta = \frac{25}{100} = \frac{1}{4} \]

### Step 6: Calculate \( 55 \sin \theta \)
Now we can find \( 55 \sin \theta \):
\[ 55 \sin \theta = 55 \times \frac{1}{4} = \frac{55}{4} = 13.75 \]

### Final Answer
The value of \( 55 \sin \theta \) is \( 13.75 \).

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