To find the $x$-component of the velocity of the arrow 3.00 s after it leaves the bowstring, we can use the following equation:

$v_{x} = v_0 \cos(\theta)$

where:

• $v_{x}$ is the $x$-component of the velocity,
• $v_0 = 20.6 \, \text{m/s}$ is the initial speed of the arrow,
• $\theta = 60.0^\circ$ is the angle at which the arrow was shot above the horizontal.

The key point here is that the horizontal velocity ($v_{x}$) remains constant throughout the motion (assuming no air resistance), so the $x$-component of the velocity doesn't change over time.

Step-by-step calculation:

$v_{x} = 20.6 \cdot \cos(60.0^\circ)$

We know that $\cos(60.0^\circ) = 0.5$, so:

$v_{x} = 20.6 \cdot 0.5 = 10.3 \, \text{m/s}$

Thus, the $x$-component of the velocity is $10.3 \, \text{m/s}$.

Let me know if you want further details or have any questions.

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Here are 5 related questions to expand on this topic:

1. What is the $y$-component of the velocity 3.00 seconds after the arrow is shot?
2. How would air resistance affect the $x$-component of the velocity over time?
3. What is the total velocity of the arrow 3.00 seconds after it was launched?
4. How can we calculate the time the arrow spends in the air before landing?
5. What would the $x$-component of velocity be if the arrow was shot at a different angle?

Tip: In projectile motion, horizontal velocity remains constant when air resistance is neglected, while vertical velocity changes due to gravity.