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Homework Help: 1 dimensional problem, from Yale online physics course.

  1. Aug 21, 2016 #1


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    Gold Member

    Hi All,

    I found a free online course Physics from Yale I thought that must be pretty good well after watching the lecture and starting with the first problem I think I found something that is incorrect. Is anybody willing to tell me that i do somehting wrong (most probably I would expect that the prof at Yale has a pretty good understang of physics).

    1. The problem statement, all variables and given/known data
    From the top of a building of height h = 100 m I throw a stone up with velocity 10 m/s. What is
    the maximum height it reaches, and when does this occur? How many seconds does it spend on
    its way down between h = 50 m and h = 0 m? What is its velocity when h = 50 m? If, while the
    stone is airborne, an earthquake opens up a hole 50 m deep in the ground, when and with what
    speed will the stone hit the bottom?

    2. Relevant equations
    1) X(t) = X0 + v0t + 1/2at2
    2) v = dX/dt = v0 * at ⇒ t = vt - v0 / a

    Now when we plug in for the equation to find t in the first equation we get equation 3 below.

    Xt - X0 = vt2 - v02 / 2a

    So far so good, I guess

    3. The attempt at a solution
    We know when we throw up the stone at 10m/s at a certain point the speed will be 0m/s due to the gravitational force thats pulling on the stone so lets find out first how much time this takes with Eq 2:

    v = dX/dt = v0 * at
    0 = 10 -9.8 * t
    t = 10 / 9.8 = 1.0 s

    Now to find the maximum hight we can use Eq 1.
    X(t) = X0 + v0t + 1/2at2

    Xmax = 100 + 10*1 - 1/2 * 9.8 * 12 = 105.1m

    So far so good regarding the Proff his answer and mine. But now the weird stuff is going to happen. the proffesor. The next question we need to know is how much time it takes on it's way down between 50m and 0m.

    We can use Eq 3 and lets call Xt - X0 = Δh so we get:
    Δh = vt2 - v02 / 2a

    Now the proff is going to calculate the two velocities at the hight asked above so he pluges in for v1 50m and v2 100m.

    v1 = -SQRT(102 + 2*9.8 * 50 = -32.9 ms
    v2 = -SQRT(102 + 2*9.8 * 100 = -45.4 m/s

    And here I dissagree: I think we should plug in 55m and 105m. This because I throwed up the stone to 105m this means the stone has more potential energy at 105m as at 100m and it will built up more speed on its way down! I think this is a crucial point of the problem but it could be that im thinking wrong offcourse!

    So I got the following answers for v1 and v2:
    v1 = -SQRT(102 + 2*9.8 * 55 = -34.3 ms
    v2 = -SQRT(102 + 2*9.8 * 105 = -46.5 m/s

    Now I can calculate the time it took by taking the difference of the 2 velocities and plug it into Eq 2

    t = -34.3- -46.5 / 9.8 = 1.2s

    the Prof had 1.3s.

    My answer is slightly different compared to the prof but that is not the point for me. It is the way how we think about what is happening physically.

    Again I think when I throw up the stone to 105m it bults up more potential energy which is converted on its way down to kinetic energy. And it can built up little more from 105m instead of 100m.

    So my big questin here is. Am I missing something here it cant be that a Prof at Yale missed such a crucial point in the problem solving and a rookie like me sees this. So in fact im think im wrong!

    And comments would be highly appreciated.

    Grtz Raymond
  2. jcsd
  3. Aug 21, 2016 #2

    Doc Al

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    Staff: Mentor

    Yes, you're wrong. It's true that the stone goes up to some height above the starting point, but that's already covered in the initial velocity. And you are told that you need to find the time between points h = 50 and h = 0, so you cannot go changing that.
  4. Aug 21, 2016 #3


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    Thanks a lot for the response.

    Unfortunately I dont understand it. So h=50 is 50m for the max 105m or 50m from the 100m? This is what I dont undertsand, sounds maybe very silly question but thats not clear to me.
  5. Aug 21, 2016 #4

    Doc Al

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    Staff: Mentor

    h is the height above the ground. Since you are told to find the time between the points where h = 50 and h = 0, those are heights you should use.

    When you use the following kinematic equation
    ##v^2 = v_0^2 + 2 a \Delta x##
    realize that the change in position must be measured from the point where you measured the initial velocity. Since you're using ##v_0 = 10##, you are measuring from h = 100.

    If instead you used the highest point of the motion, the initial velocity would be set to zero, not 10 m/s. Make sense?
  6. Aug 21, 2016 #5


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    Ahhhh offcourse EUREKA ;) yes that makes fully sense!

    If I would do it my way I would have set the initial velocity to 0m/s and I will come on the same numbers! :)

    I knew I made a wrong way of thinking somewhere, impossible that a rookie like me finds a mistake from a Prof at Yale, lol

    Thanks a lot
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