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Homework Help: Maximum height

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

    2. Relevant equations
    Vf^2 = Vi^2 +2gt

    3. The attempt at a solution
    Couldn't figure out Vi..
    25 m/s ^ 2 = Vi ^ 2 + 2 (gravity) 2/3Y(time)
  2. jcsd
  3. Jan 29, 2014 #2


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    Need a bit more. Equations too.
    What is the sign of (gravity) ?
  4. Jan 29, 2014 #3
    I meant to say garvity = g.
    Could you tell me what else I need to get the maximum height.
  5. Jan 29, 2014 #4


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    Yes I could, but PF encourages you to think a little bit too. Is g +9.81 or is it -9.81 ?

    In this kind of exercises, you generally need more than one equation.

    Uniform motion: x(t) = x(0) + v * t with v a constant. One equation.

    Uniformly accelerated motion:
    x(t) = x(0) + ....
    v(t) = v(0) + ....
    Two equations.

    You meant Vf^2 = Vi^2 +2g Y, right ?

    This equation is an energy balance, which is even more useful: 1/2 m vf^2 - 1/2 m vi^2 = mgh
    can be used to get an equation with 2/3 h and vi

    With vf = 0 at the highest point, you get another equation with h and vi.

    So I'm embarrased to say that I am contradicting myself: here you can make do with this one equation . Still have to do some solving...
  6. Jan 29, 2014 #5


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    Here's a hint: forget about the ground and the "initial" velocity.
  7. Jan 29, 2014 #6
    It's +9.8 m/s2.
    So second equation would be V(t) = V(o) + at?
    V(t) = 25m/s + 9.8 m/s2 2/3 y?
  8. Jan 30, 2014 #7


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    First, if something is travelling upwards at 25m/s, it will travel x m from that point. So, the first thing to do is to calculate x.

    The general equation for this is:

    [tex]x = \frac{v^2}{2g}[/tex]
    Here you have v = 25m/s.

    Then, in this case, the maximum height, h, is given by:

    [tex]h = 3x[/tex]
    As, x was only the last third of the trajectory.
  9. Jan 30, 2014 #8


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    Let's step back a little.
    What we have here is something that goes up and is being pulled down by the earth. It goes up with some upward speed, goes slower and slower until its vertical speed is zero and then it starts to fall down. The thing to realize is that there is a direction attached to the velocity and also to the acceleration. Since our "object" is thrown up vertically and should drop back right on the head of the dumb thrower, all the motion takes place along a single vertical line, which we often designate as the y-axis. So-called 1 dimensional linear motion with constant acceleration.

    We're not done yet with our formalizing: choosing a coordinate system to do some calculations involves not only picking a good line --vertical in this case-- as axis, but also a direction that we designate as positive. You save yourself a lot of trouble if you follow the conventional choice: up is positive, down is negative. By the time your "objects" have to fly around the world or other planets, you will have enough experience to then make the right choices, so that's for later.

    This is so important that I dedicate a fancy drawing ;-) to celebrate this first crucial step:
    (If I weren't so old, I would know how to include the picture reasonable size here instead of as an attachment. Never mind, someone will give me a tip and we don't want to get distracted now).

    As you can see, I have only drawn one half of the y axis, the positive half. As soon as we calculate a negative y somewhere, we have to check, double check and check again. It's not completely impossible (wells, for instance), but usually things travel differently above ground than below.

    All we need to do now to really get going is designate a unit vector in the positive y-direction. Something that is 1 long and points upward. In view of the wording in the OP, I propose a 1 meter long nice fat arrow standing up at the origin (the surface of the earth).
    From now on y is a vector with the dimension of length and a magnitude expressed in meters. Convention: y fat is vector, y normal is magnitude.

    Same coordinate system for velocity vy = change in y per unit time. Since we're talking linear motion (one dimension in space, but that confuses the term dimension in physics), the y subscript is nice but not essential. Different beast: v lives in the same coordinate system but has a different (physical) dimension, namely m/s in our case. I'm not allowed to shout in PF, but you can NOT add things that live in different (physical) dimensions. So when you end up adding y to v you have either forgotten a factor time somewhere or you're working so sloppily that you deserve to flunk. All in good spirit.

    (I lost the rest of this post - quite a bit) so now I first post this and then continue at leisure with the remainder.

    Attached Files:

    Last edited: Jan 30, 2014
  10. Jan 30, 2014 #9


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    Back to the problem. In our (my) new coordinate system the proper equation is V(t) = V(o) + at. Vectors with the same dimension (m/s). At a certain point you want to do some calculation with numbers. Magnitudes of vectors are non-negative numbers. But it's always good to hang on to dimensions. Somewhat double. So the number variety is V(t) = V(o) + at. Provided you take the direction into account. In our system V is up and decreases because the earth is pulling down. Two ways to do this:
    preferred way: write V(t) = V(o) - g t with g = 9.81 m/s2
    equally correct but more confusing/error prone: V(t) = V(o) + g t with g = -9.81 m/s2. Potential confusion: because non-bold suggests magnitude and that can't be negative.

    Both are correct. So is the other (stubborn, confusing, error prone, unconventional) choice to let the positive y axis point downwards. Even there v and g point in opposite directions. Never mind.

    What is definitely wrong is to fill in the numbers and ignore the directions, writing down
    V(t) = 25m/s + 9.8 m/s2 t because the vector with length 25 points in the opposite direction from the vector with length 9.8 t. (Silently I chose t = 0 at the moment of throwing).

    Let's look at another (the relevant equation):
    "Vf^2 = Vi^2 +2gt" Can't be right: v2/ is m2/s2 and g t is m/s2 * s = m/s. That's why I hinted "You meant Vf^2 = Vi^2 +2g Y, right ? "

    You had already made that step, because in your attempted solution it says
    25 m/s ^ 2 = Vi ^ 2 + 2 (gravity) 2/3Y(time) with correct dimensionality. Let's assume a typing mistake. The questions about sign of (gravity) were meant to get you on the track that the 25 m/s is slower than Vi, something that is only possible if gravity and Y have different signs.

    So: corrected version: 25 m/s ^ 2 = Vi ^ 2 - 2 (gravity) 2/3 Y(time) Vi is upwards speed at throwing time (which we silently take as zero). Interpretation: Y(time) is the positive magnitude of the maximum height. And "time" is the time the thing is at its highest point. Y(time) is the thing they want us to find. All very well, but there are three unknowns: Vi, time and Y !

    Perok is kind of short-cutting towards the answer, which is fine if it were important. My lengthy monologue is aimed at letting you understand what you are doing. If I succeed, things will be a lot easier, because then you know what needs to be done and you only have to concentrate on doing it correctly.

    We're temporarily in bad shape: one equation, three unknowns! However, we have no interest in time (the number of seconds it takes to get at this highest point). Nor in Vi (not asked for).

    We can write down something for time time (the number of seconds it takes to get at this highest point): at that moment | V | = V = 0 . And Y is Y(time):

    (0 m/s)2 = Vi 2 - 2 g Y(time)

    Subtracting the two gets rid of Vi2 . Good riddance. Left with

    (25 m/s)2 = - 2 g (2/3 Y(time) - Y(time))

    And Y(time) rolls out. time does not, but we don't need that.

    If you want to delight me, work out time and Vi. In return for some more free tuition, if ever needed. Reason I ask: you'll need stuff from the equations for uniformly accelerated linear motion and you want to master that anyway.
    Last edited: Jan 30, 2014
  11. Feb 1, 2014 #10
    625 m^2/s^2 = 2gy - 4/3 gy
    y = 96 m
    Thanks for detailed explanation.
    wonderful BvU
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