Answer:
95 m
Explanation:
To solve this problem, we need to consider the motion of the lead fishing line in two dimensions separately: horizontal and vertical. The key here is that the horizontal motion is unaffected by gravity and remains constant, while the vertical motion is influenced by gravity.
We are given:
We also know:
We will use the four kinematic equations below:
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{The 4 Kinematic Equations:}}\\\\1. \ \vec v_f=\vec v_0+\vec at\\\\2. \ \Delta \vec x=\frac{1}{2}(\vec v_f+\vec v_0)t\\\\3. \ \Delta \vec x=\vec v_0t+\frac{1}{2}\vec at^2\\\\ 4. \ \vec v_f^2=\vec v_0^2+2\vec a \Delta \vec x \end{array}\right}[/tex]
Using the vertical components and equation (3), determine the time the projectile is in the air for:
[tex]\Delta \vec y=\vec v_{0_y}t+\frac{1}{2}\vec a_yt^2\\\\\\\\\Longrightarrow -196.4 \text{ m}=(0 \text{ m/s})t+\frac{1}{2}(-9.8 \text{ m/s}^2)t^2\\\\\\\\\Longrightarrow -196.4 \text{ m}=(-4.9 \text{ m/s}^2)t^2\\\\\\\\\Longrightarrow t = \sqrt{\dfrac{-196.4 \text{ m}}{-4.9 \text{ m/s}^2}}\\\\\\\\\therefore t \approx 6.33 \text{ s}[/tex]
Thus, the projectile is airborne for approximately 6.33 seconds. We can use this information, along with the horizontal components to find the horizontal distance. Use equation (3) again:
[tex]\Delta \vec x=\vec v_{0_x}t+\frac{1}{2}\vec a_xt^2\\\\\\\\\Longrightarrow \Delta \vec x=(15 \text{ m/s})(6.33 \text{ s})+\frac{1}{2}(0 \text{ m/s}^2)(6.33 \text{ s})^2\\\\\\\\\therefore \Delta \vec x \approx \boxed{95 \text{ m}}[/tex]
Thus, the horizontal displacement of the lead fishing line before striking the water is approximately 95 meters.
