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

Explanation

To solve the problem of determining the height at which the water strikes the wall, we can break down the problem into several steps:

### Step 1: Identify the Components of the Initial Velocity
The initial velocity of the water is given as \( u = 16 \, \text{m/s} \) at an angle of \( 60^\circ \) to the horizontal. We need to find the horizontal and vertical components of this velocity.

- **Horizontal Component (\( u_x \))**:
\[ u_x = u \cos(60^\circ) = 16 \cos(60^\circ) = 16 \times \frac{1}{2} = 8 \, \text{m/s} \]

- **Vertical Component (\( u_y \))**:
\[ u_y = u \sin(60^\circ) = 16 \sin(60^\circ) = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3} \, \text{m/s} \]

### Step 2: Calculate the Time to Reach the Wall
The wall is located \( 8 \, \text{m} \) away from the hose. We can find the time taken to reach the wall using the horizontal component of the velocity.

\[ \text{Time} (t) = \frac{\text{Distance}}{\text{Velocity}} = \frac{8 \, \text{m}}{u_x} = \frac{8}{8} = 1 \, \text{s} \]

### Step 3: Calculate the Height at Which Water Strikes the Wall
Now we can calculate the height (\( h \)) at which the water strikes the wall using the vertical motion equation. The formula for vertical displacement is given by:

\[ h = u_y t - \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Substituting the values we have:
\[ h = (8\sqrt{3}) \times 1 - \frac{1}{2} \times 10 \times (1)^2 \]
\[ h = 8\sqrt{3} - 5 \]

### Step 4: Calculate the Numerical Value of \( h \)
Now we need to calculate \( 8\sqrt{3} \) and then subtract \( 5 \):

\[ \sqrt{3} \approx 1.732 \implies 8\sqrt{3} \approx 8 \times 1.732 \approx 13.856 \]
Thus,
\[ h \approx 13.856 - 5 = 8.856 \, \text{m} \]

### Conclusion
The height at which the water strikes the wall is approximately \( 8.9 \, \text{m} \).

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