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I'm struggling with a Kinematics homework question:

On a beautiful day you decide to go fishing with your younger brother Ben. Ben has not been fishing before so he doesn't know how to properly cast out his line. He puts the lure $29.1 \mathrm{cm}$ above the water before letting it drop straight down. The line accelerates until it hits the water then continues to the bottom of the lake at a constant speed. It takes the lure $6.10 \mathrm s$ to reach the bottom of the lake from when it was released by Ben.

How deep is the lake? (Assume the lure accelerates at the free fall rate of $9.8 \mathrm{m/s^2}$ until it hits the water.)

I'm really not sure what I'm doing wrong here. The steps I took were:

$$ t_f-t_i=6.10\ \mathrm s$$ $$ a = -9.8 \ \mathrm{m/s^2}$$ $$ x_f - x_i = \text{height} = (1/2)at^2 = (1/2)(-9.8)(6.10)^2 = 182.329\ \mathrm{cm}$$

Now I subtract $29.1 \mathrm{cm}$ from the height to get: $182.329 - 29.1 = 153.229\ \mathrm{cm}$

I feel like it's a really silly mistake but there's no way to check as for some reason my textbook does no examples like these.

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  • $\begingroup$ Hello there, and welcome to the Physics Stack Exchange! Homework and "check my work" questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. Please read this post on asking homework questions and this post for "check my work" questions. $\endgroup$
    – Vincent Thacker
    Commented Sep 15, 2021 at 14:22

2 Answers 2

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The motion is not uniformly accelerated throughout. From the time it is dropped to the time it touches the lake's surface, the motion is uniformly accelerated. But after touching the lake's surface till it hits the lake's bottom, the acceleration is zero. You need to separately calculate for both parts of the motion.

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  • $\begingroup$ Ahh I see what you mean. Now what I tried was to find the velocity when it hits the water like so: $$ v_f^2=2\left(-9.81\right)\left(29.1\right) $$ and then use $$ d = (1/2)(v_f + v_i)t $$ $\endgroup$
    – Melody Banks
    Commented Sep 15, 2021 at 6:18
  • $\begingroup$ Your approach is pretty indirect. And you are not consistent with your sign conventions. You have taken the downward direction to be negative for acceleration but positive for displacement. $v_f$ will turn out to be an imaginary number in your calculations. $\endgroup$
    – Mechanic
    Commented Sep 15, 2021 at 6:26
  • $\begingroup$ Oh you're absolutely right, I wasn't careful about the signs. But did I setup the equations the right way? I'm honestly just trying to figure out how to set it up $\endgroup$
    – Melody Banks
    Commented Sep 15, 2021 at 6:31
  • $\begingroup$ Yes, you did set up the right way. $\endgroup$
    – Mechanic
    Commented Sep 15, 2021 at 6:41
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Your equations assume constant acceleration the whole time. I suggest reading the question more carefully.

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  • $\begingroup$ But it says to assume $$9.8 m/s^2$$ for the free fall to the bottom of the lake. That means the acceleration should be the same the whole way down no? $\endgroup$
    – Melody Banks
    Commented Sep 15, 2021 at 5:59
  • $\begingroup$ Oh hang on no, I misread $\endgroup$
    – Melody Banks
    Commented Sep 15, 2021 at 5:59

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