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1-D Projectile Motion: Throwing a Rock

  1. Jan 26, 2017 #1
    1. The problem statement, all variables and given/known data

    You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand? Ignore air resistance.

    2. Relevant equations

    y= y0+v0t-1/2*gt2
    v= v0 -1/2*gt.

    3. The attempt at a solution
    I used v= v0 -1/2*gt to arrive at the velocity that the rock was going at the moment it was passing the hand. My answer was v= 0- 1/2*(9.8m/s2), which gave me 24.5m/s. I have an idea of where to go from here. I know I need to use v2 − 2g(y − y 0).

    My issue is that I don't want to just memorize the equation. I need help deriving it from the two basic equations: y= y0+v0t-1/2*gt2 and v= v0 -1/2*gt. Once I do that, I believe I can finish the problem on my own.
     
  2. jcsd
  3. Jan 26, 2017 #2

    PeroK

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    So, you want to derive?

    ##v^2 - u^2 = 2as##
     
  4. Jan 26, 2017 #3
    I think so, although I'm not sure what the s and u are for.
     
  5. Jan 26, 2017 #4

    PeroK

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    ##u## is the initial velocity and ##s## is the displacement. I'm not sure why ##s## is used, but it's fairly standard.
     
  6. Jan 26, 2017 #5

    PeroK

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    Here's my favourite derivation, as it reflects the way I think about kinematic problems:

    ##v^2-u^2 = (v-u)(v+u)##

    Now:

    ##v-u = at##

    Because the difference between the final and initial velocities is just the acceleration multiplied by the time.

    And:

    ##\frac{v+u}{2} = v_{avg} \ \ ## is the average velocity (when you have constant acceleration). Hence:

    ##v+u = 2v_{avg}##

    And

    ##s = v_{avg}t \ \ ## because the displacement is simply the average velocity multiplied by the time.

    So, putting this all together:

    ##(v-u)(v+u) = at(v+u) = 2atv_{avg} = 2as##
     
  7. Jan 26, 2017 #6

    PeroK

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    You need to check where you got this second equation from.
     
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