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Finding the Value of Gravity Given Height and Velocity

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    If a ball is dropped from a height (H) its velocity will increase until it hits the ground (assuming that aerodynamic drag due to the air is negligible). During its fall, its initial potential energy is converted into kinetic energy. If the ball is dropped from a height of 880cm, and the impact velocity is 43 feet per second, determine the value of gravity in units of meters per second squared.

    2. Relevant equations
    Potential Energy = (mass)(acc of gravity)(height)
    Kinetic Energy = 1.2(mass)(final velocity2 - initial velocity2)
    =1.2(mass)(velocity)2 ///[FOR NONROTATING BODY]
    Work = Force * Distance

    3. The attempt at a solution
    I am really at a loss for this one. If I plug in impact velocity to find Kinetic Energy (which I don't think I can do since it gave me the impact velocity) I get the following:

    KE = 1/2(m)(43f/sec)


    If I plug in what I know for Potential Energy I get the following:

    PE = m(g)880cm


    Also once I convert the 2 I get:

    H = 8.8m
    iV = 13.72m/s


    I do not see how I can solve anything here since I do not have KE or mass. I feel like I need another value before I can do anything, and I have looked extensively in my textbook for examples like this but I have found none that were answered. I also can't get the average velocity since I only know the impact velocity (I would assume it would be (43f/sec)/2), but even if I had that what would be the point? I have no idea where to start, in my class we have went over only the formulas posted and I can't think of a way to solve this. Thanks in advance.
     
  2. jcsd
  3. Sep 28, 2016 #2

    gneill

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    Staff: Mentor

    This question is probably from a course unit on conservation of energy or the work-energy theorem.

    What do you know about the relationship between the change in KE and the change in PE for a projectile?
     
  4. Sep 28, 2016 #3
    Nothing really. This questions is for an introductory Engineering Class surprisingly (Engineering Problem Solving ENGR1300). I know that potential energy is weight (mass * gravity) * height and that Kinetic Energy is 1/2(mass)(velocity)^2. I also know that energy is Force * distance(or height). This is the extent, however. Most of the class is spent converting units and plugging them into given equations, however with the equations we were given here I do not see how that is possible. Below is a full list of the equations mentioned in the chapter we are on (although we have only used 3 so far: PE, KE, Q):

    Work
    W = F*deltaX
    Potential Energy
    PE = mg(deltaH)
    Kinetic Energy, transitional
    KET=1/2m(v2f - v2i)
    Kinetic Energy, rotational
    KER = 1/2(w2-w2)
    Kinetic Energy, total
    KE = KET + KER
    Thermal Energy
    Q = mCp(deltaT)
     
  5. Sep 28, 2016 #4

    jbriggs444

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    Science Advisor

    You also know that, in this case, the potential energy before the drop is equal to the kinetic energy at the end of the drop. That gives you an equation.
     
  6. Sep 28, 2016 #5
    So I can set:

    m(g)8m = 1/2m(13.72m/s)

    But how does that help me solve for g?
     
  7. Sep 28, 2016 #6

    gneill

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    Staff: Mentor

    You might want to choose a different variable name for mass to distinguish it from 'm' for meters if you're going to mix variables and numericals. maybe use M instead.

    What cancels out in your equation? can you g?
     
  8. Sep 28, 2016 #7
    Ah I see, so these are the steps to find the equation for g:

    mgH = 1/2m(v)2
    g = (1/2m(v)2)/mH
    g = 1/2(v)2/H ///because mass cancels out

    Then I plug the info in

    g = (1/2(13.72m/s)2) / 8.8m
    g = (1/2(188.24m2/s2) / 8.8m
    g = (94.12m2/s2) / 8.8m
    g = 10.7m/s2

    Awesome, this makes sense when I think logically about potential energy being the same as the kinetic once it stops. I think what confused me the most was not understanding if the velocity given at the stop was the one I was supposed to use. Thanks a bunch guys this one was a real brain-kicker. I take physics next semester so hopefully the information I get in this class will give me a nice little kickstart.
     
  9. Sep 28, 2016 #8

    gneill

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    Staff: Mentor

    Your method is correct. You might want to check your conversion from ft/sec to m/sec for the velocity.
     
  10. Sep 28, 2016 #9
    Thanks and yeah I caught that when entering my answer luckily.
     
  11. Sep 28, 2016 #10
    There are many ways to solve this, which is one of the beauties of physics. Let's use basic kinematic equations for the definition of what acceleration and velocity are, then use the Law of Conversation of Energy and the formulas for energy.

    d=8.80m
    final v = 13.12 metres/s


    1)
    we know that velocity is the change in velocity/s, thus a = Δv / t

    The average velocity = Δd / Δt = (v + u) / 2

    solving both for t

    v - u / a = Δd / (v+u / 2)

    simplifying it and solving for a

    ( v2 - u2 ) / 2 d = a
    considering u = 0 then v^2 / 2 d = a = g

    2)
    because of conservation of energy, the work done to bring an object to a certain height = potential energy,
    Work = F d
    F = ma = mg
    therefore Ep = mgd

    the potential energy at d height = kinetic energy on impact,
    mgd = 1/2mv2

    thus v2=2gd

    g= v2/ 2d
     
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