Given:
- Angle of projection (θ) = 60°
- Initial velocity (u) = 16 m/s
- Horizontal distance (x) = 8 m

We need to find the height at which the water strikes the wall.

Approach:
1. Split the initial velocity into horizontal and vertical components.
2. Use the equations of motion to find the time of flight.
3. Use the time of flight to find the maximum height reached by the water stream.
4. Use the maximum height and the horizontal distance to find the height at which the water strikes the wall.

Solution:

Step 1: Splitting the initial velocity
The initial velocity can be split into horizontal (ux) and vertical (uy) components using trigonometry:

ux = u * cos(θ)
uy = u * sin(θ)

Substituting the given values:
ux = 16 * cos(60°)
ux = 16 * 0.5
ux = 8 m/s

uy = 16 * sin(60°)
uy = 16 * √3/2
uy = 8√3 m/s

Step 2: Finding the time of flight
The time of flight (t) can be calculated using the equation:

t = 2 * uy / g

Where g is the acceleration due to gravity (g = 9.8 m/s²).

Substituting the values:
t = 2 * (8√3) / 9.8
t = (16√3) / 9.8
t ≈ 1.83 s

Step 3: Finding the maximum height
The maximum height (H) reached by the water stream can be calculated using the equation:

H = uy² / (2 * g)

Substituting the values:
H = (8√3)² / (2 * 9.8)
H = 192 / 19.6
H ≈ 9.8 m

Step 4: Finding the height at which the water strikes the wall
Using the horizontal distance (x) and the time of flight (t), we can find the height at which the water strikes the wall.

The horizontal distance can be calculated using the equation:

x = ux * t

Substituting the values:
8 = 8 * 1.83
8 = 14.64 m

Therefore, the water strikes the wall at a height of 14.64 m.