To solve this, we can use kinematic equations. We are given the maximum height reached by the water squirt, which is $45.9 \, \text{cm} = 0.459 \, \text{m}$. Our goal is to find the initial speed $v_0$ of the water when it leaves the syringe.

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Step-by-Step Solution

1. Kinematic equation for vertical motion:
   Since the water reaches its peak height when the velocity becomes zero, the following equation applies:

$v^2 = v_0^2 - 2 g h$

Where:

• $v = 0 \, \text{m/s}$ (the velocity at the maximum height)
• $v_0$ is the initial velocity (what we need to find)
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)
• $h = 0.459 \, \text{m}$ (maximum height reached)

2. Rearrange to solve for $v_0$:

$0 = v_0^2 - 2 g h \quad \Rightarrow \quad v_0^2 = 2 g h$

3. Substitute known values:

$v_0^2 = 2 (9.81) (0.459)$

$v_0^2 = 9.00438$

$v_0 = \sqrt{9.00438} \approx 3.0 \, \text{m/s}$

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Final Answer

The water emerges from the syringe with an initial speed of approximately:

$\boxed{3.0 \, \text{m/s}}$

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Do you want more details or have further questions?

Here are 5 related questions to explore further:

1. How much time does the water take to reach the maximum height?
2. What would happen to the speed if the syringe were squeezed twice as hard?
3. How does air resistance affect the trajectory in real-life scenarios?
4. If the syringe were pointed at a 45-degree angle, how far would the water travel horizontally?
5. What is the potential energy of the water at the peak height?

Tip: In vertical motion problems, always set the velocity at the maximum height to zero to simplify your calculations.