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Help with kinematics equations (d = v t and the projectile one)

  1. Dec 3, 2007 #1

    1. i would like to ask a physics question concerning two basic kinematics equations: d = v t and the projectile equation d = v t + 1/2 a t(squared). i was wondering on how exactly one would take these two equations and (using ONLY the symbols) prove that they should be equal. what i mean is, if these two equations were applied to a situation whereby two balls (one dropped and one projected in the horizontal direction) are launched at the same moment and from the same height, how would one prove that the equations are equal thus, showing that the two balls should, in fact, land on the ground at the same time since only the vertical path of motion is accounted for. note: one must remember that gravity acts on both balls at the same rate.

    2. so to rephrase: how would one take d = v t (for the dropping ball) and d = v t + 1/2 a t(squared) (which is the "y" component for the projected ball in the vertical direction) and make them equal to each other, showing that t = t OR d = d. use only the symbols!

    3. this is what i think i should do/ what i did so far: take d = v t and sub it into the other equation and solving for t... but i got stumped because i had trouble continuing. also, i tried it another way, making making d = v t and re-arranging for t and subbing that into the second equation. are these the right ideas? and will i get t = t OR d = d in the end? if not, then what's the idea that i should be looking for?

    help is very much appreciated. thanks in advance.

  2. jcsd
  3. Dec 3, 2007 #2

    Well, firstly, there's only one equation-- and it goes something like:

    [tex] x = x_0 + vt + \frac{1}{2}at^2 [/tex]

    where x is the new position,
    [itex]x_0[/itex] = the initial position,
    v = velocity,
    a = acceleration, and
    t = time.

    So when you're talking about two equations, d = vt and d = v t + 1/2 a t(squared), you're really talking about one equation. The first one (d = vt) assumes that [itex]x_0[/itex] and a are both = 0 (start from zero, and no acceleration), whereas the second equation only assumes that you start from zero.

    Now, you're trying to figure out motion in two dimensions. In that case, you have motion in both the x direction and the y direction, and so you have two equations of motion:

    [tex] x = x_0 + v_xt + \frac{1}{2}a_xt^2 [/tex]

    [tex] y = y_0 + v_yt + \frac{1}{2}a_yt^2 [/tex]

    Note that motion in either direction is only affected by the velocity or acceleration that occurs in that direction-- I'm not sure if you've gotten into vectors yet, but the implication of this is that an acceleration that, say, only acts in the y direction (gravity, for instance) will have no impact on the x coordinate.

    Hope this helps.
  4. Dec 3, 2007 #3
    many thanks to your efforts in trying to aid me and i understand all that you have written above, but it does not fully answer the question that i was orginally implying. i'll take it as my fault.

    however, let me rephrase and re-explain the situation:

    - i have made this apparatus which can do two things at once, drop one ball and project another
    - at the exact same time one ball is dropped (straight down onto the ground), the other ball is projected (note that both balls are of the same height in their initial position at rest)
    - now, the purpose of this device is to prove that the rate at which an object falls is not dependant on its mass because acceleration due to gravity acts on all objects at the same rate
    - knowing this, i ask: how would you show this situation using two equations (one for the dropping ball and one for the projected one) to show that, if launched from the same height -- which means that vertical component d is constant for both balls -- the balls will both hit the ground at the same time (where t of dropped ball = t of projected ball)??
  5. Dec 3, 2007 #4


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    I don't know why one ball has to be projected since you could prove the same point by just dropping them both at the same time. I think the thing you're trying to get to is the equivalence principle. What you'll need to do is use [itex] F=ma[/itex] and [itex] F=G\frac{mM}{r^2} [/itex] in which case the masses cancel as we assume the inertial mass from Newton's second law is the same as the mass used in Newton's law of universal gravitation.
  6. Dec 3, 2007 #5
    the reason one ball is projected and one isn't is because i wanted to prove that whether a ball drops straight down or is projected, they will hit the ground at the same time. this being said, my point of projecting one of the balls is to also show that the horizontal direction that it travels in is not related to its vertical motion (which happens to be travelling at the same amount of distance as the one falling straight down for every equal time interval that passes).

    and thanks for your input also, but i would like to avoid the universal gravitation laws and Newton altogether. for this case, i would rather stick to kinematics equations. thanks.
  7. Dec 3, 2007 #6
    I know. I'm not going to do your work for you :P I've given you the tools you need to solve this problem above, along with a huge hint as to what you need to do.

    Have you drawn a picture of what you're trying to do? You say,

    I don't know what you mean by "the vertical component of d is constant for both balls", but you here propose a way to proceed-- to prove that t of dropped ball = t of projected ball. What are the equations of motion for each ball?
  8. Dec 3, 2007 #7
    well, since you need to consider acceleration for both balls, then the equation d = v t + 1/2 a t(squared) should apply to both also. therefore, since d = d:

    v t + 1/2 a t(squared) = v t + 1/2 a t(squared)
    t(v + 1/2at) = t(v + 1/2at)

    after factoring the ts the "(v + 1/2at)" for both sides can be eliminated thus leaving t = t.

    is this correct?

    dotman: with all due respect, i've needn't to draw a picture as i have already built the device
    Last edited: Dec 3, 2007
  9. Dec 3, 2007 #8
    the reason that the "(v + 1/2at)" can simply cancel out is because both balls initially start at rest (where v = 0) and both balls experience the same rate of acceleration when they are "let loose". does it then make sense to cancel out "(v + 1/2at)" from both sides of the factored equation (from previous post)? but what i am not sure about is the fact that one t still remains within the brackets.. does this mean that one can just make an assumption that it is equal and thus, cancel it out? or am i wrong about this?
  10. Dec 3, 2007 #9
    Well, it's correct insofar as you haven't done anything wrong mathematically, but you haven't proven anything really-- ie, setting the same equation equal to itself, and then dividing out all of the variables, simply leaves 1=1. Which is well and good, but doesn't really show anything. What is d supposed to represent?

    All due respect, but you do. The point of drawing a free body diagram is to help you understand the forces involved, and especially their directions, which is of paramount importance here.

    I think you're on the right path, but you need to label which equation of motion is for which ball, and then demonstrate why they are equal, correct? I showed you the equations of motion for two dimensions, one for the x direction, and one for y. Have you tried writing these equations for each ball? From another angle, can you explain in words why or why not both balls hit the ground at the same time?
  11. Dec 3, 2007 #10
    dotman, when i set them equal to each other i meant to have one equation representing the falling ball and one representing the projected ball. i didnt use subscripts to label them because i have not any clue on how to do that. but the leftside represents the falling ball and the right side is the projected ball. d represents the vertical distance each ball is travelling and since they are released from the same height aobve the ground, the d for both balls and thus, both equations, are equal.

    i am trying to avoid the use of forces and dynamics in this situation for a good reason. the question specifically wants only the use of kinematics without the use of force, mass, or any other dynamic/ Newton-related theories and/or laws.

    and as i said before, only the y component of the projected ball needs to be taken into account.

    yes, the balls hit the ground at the same time if released from the same height at the exact same moment because both initially start at rest and must travel the same distance with the same gravitational force acting on them. therefore, they will make contact with the ground at the same time.
  12. Dec 3, 2007 #11
    This is what you need to, but have not, proven. Let me show you your equation with subscripts:

    [tex]v_{falling}t + \frac{1}{2}a_{falling}t^2 = v_{projected}t + \frac{1}{2}a_{projected}t^2[/tex]

    You can't divide these out unless you can show that [itex]v_{falling} = v_{projected}[/itex], etc. etc. Why should I, for instance, believe that gravity exacts the same effect on the projected ball as the falling one? Afterall, the projected ball is moving forward, surely that must have some effect! Can you prove me wrong?

    This argument will be difficult (impossible?) without the concept of vectors.
  13. Dec 3, 2007 #12
    v falling and v projected are, however, the same. as mentioned before, they start from rest. therefore v=0 for both balls and this vfalling t and vprojected t can be automatically cut out.

    this topic is, by far, getting somewhat too complicated and moving away from the question that i would like to have answered. all i simply ask for is to show that the projected ball and dropped ball both land on the ground after travelling the same VERTICAL distance and in the same amount of time.

    take into account the following:

    "The key point is that the fall of a ball from a particular height to the ground takes the same time whether or not it also has horizontal motion."

    "Different masses fall in the same time, with the same vertical acceleration."

    and also consider the following diagrams:

    the first half shows the path of the dropped ball. it travels in increasing intervals of distances in each equal time interval.
    the second half shows the path of the projected ball. as shown, it also travels the same vertical distance intervals in each equal time interval as with the dropped one.

    now is there a way to make the equations of each to show that t = t, knowing that d = d (d which is the vertical distance travelled)?
  14. Dec 3, 2007 #13
    Well, if you want to prove this mathematically, you can't just 'state' that they're the same without some proof, and expect anyone to buy it. And stating that they start from rest proves nothing with regard to [itex] v_{falling}t = v_{projected}t[/itex], since you haven't shown that the acceleration is equal.

    I've given you the necessary equations to do this.

    Seriously, how can you expect to prove that:

    Without showing anything regarding the x component of the motion, or even suggesting why the horizontal motion is irrelevant? This is the whole point of the exercise!

    Perhaps you should focus from a different angle-- why is the horizontal motion irrelevant?
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