A hose lying on the ground shoots a stream of water upward at an angle of `60^@` to the horizontal with the velocity of `16 m s^-1`. The height at which the water strikes the wall `8 m` away is.

Options

A. A. 8.9 m
B. B. 10.9 m
C. C. 12.9 m
D. D. 6.9 m

Explanation

To solve the problem of finding the height at which the water strikes the wall, we can break the motion into horizontal and vertical components. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial velocity (u) = 16 m/s
- Angle of projection (θ) = 60°
- Horizontal distance to the wall (x) = 8 m

### Step 2: Calculate the horizontal and vertical components of the initial velocity
The horizontal component of the initial velocity (u_x) can be calculated using the formula:
\[ u_x = u \cdot \cos(\theta) \]
The vertical component of the initial velocity (u_y) can be calculated using the formula:
\[ u_y = u \cdot \sin(\theta) \]

Substituting the values:
- \( u_x = 16 \cdot \cos(60°) = 16 \cdot \frac{1}{2} = 8 \, \text{m/s} \)
- \( u_y = 16 \cdot \sin(60°) = 16 \cdot \frac{\sqrt{3}}{2} = 8\sqrt{3} \, \text{m/s} \)

### Step 3: Calculate the time taken to reach the wall
The time (t) taken to reach the wall can be calculated using the horizontal motion equation:
\[ t = \frac{x}{u_x} \]
Substituting the values:
\[ t = \frac{8 \, \text{m}}{8 \, \text{m/s}} = 1 \, \text{s} \]

### Step 4: Calculate the vertical height at the time of impact
The vertical height (h) can be calculated using the vertical motion equation:
\[ h = u_y \cdot t - \frac{1}{2} g t^2 \]
Where g = 9.81 m/s² (acceleration due to gravity).

Substituting the values:
\[ h = (8\sqrt{3}) \cdot (1) - \frac{1}{2} \cdot 9.81 \cdot (1)^2 \]
\[ h = 8\sqrt{3} - 4.905 \]

### Step 5: Calculate the numerical value
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ h \approx 8 \cdot 1.732 - 4.905 \]
\[ h \approx 13.856 - 4.905 \]
\[ h \approx 8.951 \, \text{m} \]

### Final Answer
The height at which the water strikes the wall is approximately **8.95 m**.

---