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To understand projectile motion, you need to break down the initial velocity into two components: horizontal and vertical. This helps analyze the trajectory separately along each axis, making complex motions simpler to study.

We start with the hose that propels water at an angle of \(60^{\circ}\) above the horizontal with a speed of \(16\, \text{ms}^{-1}\). Here’s how we determine the components:

• The horizontal component, \(v_{x}\), is found using \(v_{x} = v \cos(\theta)\). Given \(v = 16\, \text{ms}^{-1}\) and \(\theta = 60^{\circ}\), this becomes \(v_{x} = 16 \times \frac{1}{2} = 8\, \text{ms}^{-1}\).
• The vertical component, \(v_{y}\), is obtained by \(v_{y} = v \sin(\theta)\). Here, \(v_{y} = 16 \times \frac{\sqrt{3}}{2} \approx 13.86\, \text{ms}^{-1}\).

By resolving these velocity components, you simplify analyzing each direction independently, setting the stage for calculating further details.

The time of flight is the duration it takes for the projectile (in this case, the water stream) to reach a specified horizontal distance. Understanding this element is crucial as it ties together distance and horizontal velocity calculations.

Here, we want to find how long it takes the stream of water to hit the wall \(8\,\text{m}\) away. Using the formula for horizontal motion \(x = v_{x} t\), where \(x\) is the distance to the wall, and \(v_{x}\) is the horizontal velocity: \[t = \frac{x}{v_{x}} = \frac{8}{8} = 1 \text{s}\] After calculating, we find that it takes \(1\,\text{s}\) for the water to reach the wall. Understanding the time of flight reveals not just the relationship between distance and velocity but also helps in predicting when the water will reach particular heights.

Vertical motion within projectile motion describes how the water stream moves under the influence of gravity. Gravity only acts vertically, hence it will decelerate the water as it travels upwards and then accelerate it downwards.

To find out the height at which the water strikes the wall, use the equation for vertical motion: \[y = v_{y} t - \frac{1}{2} g t^2\] Here \(v_{y}\) is the initial vertical velocity component, \(t\) is time, and \(g = 9.8\,\text{ms}^{-2}\) is acceleration due to gravity. Substitute the known values \(v_y \approx 13.86\, \text{ms}^{-1}\), \(t = 1\,\text{s}\):

• Calculate \(y = 13.86 \times 1 - \frac{1}{2} \times 9.8 \times (1)^2 = 13.86 - 4.9 = 8.96\,\text{m}\).

Upon rounding, we find the height to be approximately \(8.9\,\text{m}\). This demonstrates how gravitational pull affects the vertical displacement of the projectile, showing the behavior of objects under free fall conditions.