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 `50 sin theta? (g = 10 ms^(-2))`

Answer

To solve the problem, we will use the equations of motion to find the value of \(50 \sin \theta\).

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

### Step 2: Use the vertical motion equation
Since the particle clears a wall of height \(5 \, \text{m}\) after \(2 \, \text{s}\), we can use the second equation of motion for vertical displacement:
\[ S_y = u_y t + \frac{1}{2} a_y t^2 \]
Where:
- \(S_y = 5 \, \text{m}\) (the height of the wall)
- \(u_y = 50 \sin \theta\) (the vertical component of the initial velocity)
- \(t = 2 \, \text{s}\) (time taken to reach the wall)
- \(a_y = -g = -10 \, \text{m/s}^2\) (acceleration due to gravity, acting downwards)

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation gives:
\[ 5 = (50 \sin \theta)(2) + \frac{1}{2}(-10)(2^2) \]

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

### Step 5: Solve for \(100 \sin \theta\)
Rearranging the equation to isolate \(100 \sin \theta\):
\[ 100 \sin \theta = 5 + 20 \]
\[ 100 \sin \theta = 25 \]

### Step 6: Find \(50 \sin \theta\)
To find \(50 \sin \theta\), we divide both sides by 2:
\[ 50 \sin \theta = \frac{25}{2} = 12.5 \]

### Final Answer
Thus, the value of \(50 \sin \theta\) is \(12.5\).

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