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Velocity dropped and thrown meet at what height

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose you drop a ball from rest out of a window that is at height H above the ground, and suppose the ball hits the ground with speed S.
    At the instant that you release the ball, your buddy throws a ball straight up from the ground level with the same speed S.
    Show that the balls pass each other at a height of 3/4H
    What we know: for person dropping the ball
    ##y_o = h##
    ##y_1=h##
    ##a_y=-9.81m/s^2##
    ##v_0=0m/s##
    ##v_y = S##
    Person launching the ball up:
    ##y_o = 0##
    ##y_1= h##
    ##a_y=-9.81m/s^2##
    ##v_0= S##
    ##v_y = 0m/s##

    2. Relevant equations
    ##v_y = v_0 + a_yt##
    ##y=y_0+v_0t+1/2a_yt^2##
    ##v^2_y = v^2_0+2a_y(y-y_0)##
    ##y-y_0=1/2(v_0+v_x)t##
    ## v^2_y-v^2_0=2a_y(y-y_0) ##

    3. The attempt at a solution
    So my thought process is I need to use a kinematic equation to solve for a value lets say t, for both people, one dropping the ball and one throwing the ball.
    Eq.4 ## y-y_0=1/2(v_0+v_x)t ## -> ## t=2(y-y_0)/(v_0+v_y) ##
    so now that we have an equation for a function of time, my thought process is that the time when the person throws the ball the height should be 3/4H. I'm not sure if I can do this, but I'm hoping I can set v_y = 0, I think I can use it for the person dropping the ball, but not for the person throwing the ball, but I got stuck here if I didn't.
    ##t=2(3/4H)/(S+0) ## -> t = 3H/2S
    Then for the person throwing the ball down
    ## t= 1/2(1/4H)/0+S ## -> t=H/2S
    then when you set them equal to each other you get H/2s = 3H/2s -> 3
    I'm hoping in some way shape or form that is still a t, then to test my answer I plug t=3 into equation 4 again.
    ##dropping: y-y_0= 1/2(v_0+v_y)t## -> ## 1/4H= 1/2(0+S)3 ## -> ##H=6S##
    ##throwing: y-y_0 = 1/2(v_0+v_y)t ## -> ## 3/4H=1/2(S+0)3 ## -> ##H=2S##
    Set them equal to each other to get 3 again. I think at this point its a huge stretch, and it's probably just an algebra reason why I got 3 again. I'm not sure how else to approach this.
     
  2. jcsd
  3. Feb 26, 2015 #2

    Suraj M

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    Honestly, OP I'm a bit confused by your attempt, any way i do have a better and a way better(easier) method, if your interested.
    see you can find the time taken by the dropped ball to get to a height, say x from the ground. (##t_1##)
    Also you can calculate the time required for the ball thrown from below to reach a height x from the ground ##(t_2)##. If you use the 2nd equation of motion here, it will get a bit complicated, so what i propose is, find the time for the ball to reach the top ##t_{total}## then subtracting ##t_1## from this you'll get your ##t_2##.(why?)
    so equate them and solve for x.
    I'm sorry if i eloborated too much, i'm not suppose to give you the total answer but this one requires some steps.
    Post your findings.
     
  4. Feb 26, 2015 #3

    BvU

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    Hello Mathias, welcome to PF :smile:

    You've done your best to approach this systematically and sorted out things nicely. I suppose the first ##y_1=h## is a typo and you mean ##y_1 = 0## ?
    S is an unknown that can be found using your first two equations (ball falling). If you work them out you get the third equation: it gives ##S^2##. You also get ##t_{total}##, the time to fall from h to 0.

    Why? It clearly says both start at the same time !
    Is a total mystery to me. What does that mean ? H = 3 ?


    A good thing to do is make a sketch of h versus time for both trajectories. The point where the balls pass each other is where the two lines cross ! At that point in time (i.e. at that value for t) you have ##y_{down} = y_{up}##.

    Since you have expressions for both as a function of t, this should solve nicely. From what is asked, it is clear that the only thing that doesn't fall out is h.

    If you do it straightforward, you have to solve a quadratic equation in t. But if you look at the sketch, you can argue that ##t_{pass} = {1\over 2} t_{total}## and all of a sudden it's really easy...
     
    Last edited: Feb 26, 2015
  5. Feb 26, 2015 #4

    Suraj M

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    Why the sketch, i mean it'll work in this case but other cases, a bit complicated, and also i doubt we can draw sketches in an examination(from my experience :oldcry:) no offence.
    can't we just find the time for the ball thrown to reach the top = time to reach height ##h## - time for the dropped ball to reach the point where they meet
    so no quadratic equation, 2 step simplification! and theres your answer.
     
  6. Feb 26, 2015 #5

    BvU

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    Dear Suraj,

    My advice to draw the sketch was adressed to Mathias. It will help to get an understanding of the case, which in turn, later on, will enable to handle more complicated cases in an abstract manner.

    In all my exams I always have been able to use scratch paper to make all the sketches my heart desired. Not for handing in, but to help find a solution. I am certain that if Mathias can consult PF, he can also make a sketch.
     
  7. Feb 26, 2015 #6

    Suraj M

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    Oh ok sorry for questioning.
     
  8. Feb 26, 2015 #7
    aDy4q2Y.png
    Ill first convert all the kinematics into terms of t
    1. ## v_y=v_0+a_yt## -> ##t= (v_y-v_0)/a_y ##
    2. ## y-y_0=v_0t+1/2(a_yt^2) ## -> ## 1/2(a_yt^2) + v_0t-y+y_0 ##
    3. ## y-y_0=1/2(v_0+v_y)t ## -> ## t = 2(y-y_0)/(v_0+v_y) ##

    @byu I'll try your way now, and I drew a picture of how I think its supposed to be found, I just can't seem to get to a path where I can solve it.
    To be honest, I'm not really confident in doing a quadratic but here is where I got.
    2. ## y-y_0=v_0t+1/2(a_yt^2) ## -> ## 1/2(a_yt^2) + v_0t-y+y_0 ##
    from reference of person throwing the ball
    ##t= (-b+/- \sqrt((v_0)^2-4(.5a_y)(-h)))/2(.5a_y) ##
    ##t = (S +/- \sqrt((S^2+2a_yh)))/a_y ##
    ##t =S +/- \sqrt(S^2+2(9.81)))/9.81## I'm not even really sure how to even proceed anymore
    from reference of person dropping the ball:
    ##t= (-b+/- \sqrt((v_0)^2-4(.5a_y)(-h)))/2(.5a_y)##
    ##t = (0 +/- \sqrt((0+2a_yh)))/a_y##
    ##t = 0 +/- \sqrt(0+2(9.81)h))/9.81##
    ##t = +/- \sqrt(19.62h)/9.81## -> I think this breaks an algebraic rule but
    ##t =+/- 4.429\sqrt(h)/9.81##
    ##t =+-\sqrt(h)/.45## that's really the farthest I got for this one.

    @suraj I'll try it the way I think you described it.
    using equation 3 to get in terms of t
    t=2(y-y_0)/(v_y+v_0) -> replace variables with given in reference of person who drops the ball (##t_t##) ##t_t= 2(0-H)/(S+0)## -> ##t_t = -2H/S##
    now If i think correctly this is the total time, with respect to H so the whole length. Next we should try putting in 3/4H.
    let t_f = time at 3H/4, still with respect to the person dropping the ball. ##t= -2(3H/4)/S ## -> ##t_f = -3H/2S ##.
    Now we subtract time at 3H/4 (##t_f##) from total time (##t_t##) to get,
    let ##t_u## = time of unknown: ##t_t-t_f## -> ##-2H/S - (-3H/2S) -> -2H/S + 3H/2S ## -l-> ## 2/2 *(-2H/S) -> -4H/2S ## -l-> ##-4H/2S + 3H/2S## -> ## -H/2S = t_u ##
    Then not quite sure what you wanted me to do after that. When I did another attempt like this earlier but the formula was nonsense, I said something like that ##v_f-v_i/t_f-t_i = h ## or something, and got that they where inversely proportional when I plugged in 3H/4, H/4. But I'm fairly certain that formula which I used was not valid.
     
    Last edited: Feb 26, 2015
  9. Feb 26, 2015 #8

    haruspex

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    BvU was recommending looking at a sketch and noticing something from the symmetry, namely, where they are in relation to each other at the half time mark.
     
  10. Feb 26, 2015 #9
    So if I understand you correctly t = 1/2 at 3H/4 for both, dropped and thrown.
    However I am conflicted with that because the dropped ball just goes straight to the ground, But for the ball that is thrown it get to the max height (H) then falls down again. So t total of thrown would = 2, compared to the other, so if you then set them equal to each other that's you you would get dropped: ##t_d##, thrown: ##t_r## ##2t_d = t_r ## ## t_d/t_r=1/2## is where they would meet then, and at that time 1/2 is where the height 3H/4, not sure I understand that now but I'll play with that idea. That then brought me to an idea of something we did in calculus when measuring are area under the curve, we had two curves which would cross as a point (3H/4) then would subtract the 1 curve from the other. If those curves where velocity time curves then if area under the curve is an integral then the integral of the velocity curve would yield total x position, or just current x position.
     
    Last edited: Feb 26, 2015
  11. Feb 26, 2015 #10

    haruspex

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    No, we're not saying it's necessarily 3H/4 (yet). Have you drawn a height versus time diagram showing both trajectories? What does it look like? Do you see a symmetry?
    Yes, but we don't care what happens after the thrown ball reaches maximum height.
     
  12. Feb 26, 2015 #11
    Zz6ZX0Q.png
    I imagine it would look something like that. Where red is dropped, and purple is thrown. But this is just going off of the idea of what the effects of gravity would be. I guess what I would discern then is that the functions would be (inversely proportional)? I'm not sure if that's the right word. I suppose from my graph one could say that 3H/4 = t/2 but I wouldn't know how to discern that myself without the maths in front of me that would suggest something like that, if this is the case do you know what I could search for to research it?
     
  13. Feb 27, 2015 #12

    haruspex

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    Yes, the diagram is right, but don't rush into 3H/4 - that's what we have to prove. What you can say from the symmetry of the picture is that they cross over at time t/2. Now all you have to do is calculate what fraction of the height the falling ball covers in half the time. That should be easy from the SUVAT equations.
     
  14. Feb 27, 2015 #13

    BvU

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    Hello Mathias,

    You really scared me a bit with post #7. We had a fairly simple problem with a height h and not much more to come out and then I see an overwhelming amount of expressions with numbers, square roots and ##\pm##'s. "Why make things difficult, when with a little effort they can be made nearly impossible" a senior colleague of mine used to say -- long ago.

    As I said: most things -- all except h -- are supposed to fall out of the expressions. And h is a given (but in symbolic form, as a kind of parameter). So no numbers, no ##\pm##, just symbols and perhaps here and there a square or a factor ##1\over 2##.

    Let's rewrite post #1 in two parts: one down, one up.
    At this point I would propose to make things simple, find S and the time needed to reach the ground (##\tau##), expressed in symbolic form in a minimum of knowns:

    $$ \quad (1) \rightarrow S = -g\tau\\
    \quad (2) \rightarrow 0 = h - {1\over 2} g\tau^2\quad\Rightarrow S^2 = 2gh\quad (3)
    $$
    the minus sign in S means the ball is dropping. We only need ## |S| = \sqrt{2gh}##.
    And the trajectory of the falling ball can be described with ## y = h - {1\over 2} gt^2##.

    In other words: ## y = St - {1\over 2} gt^2##.

    When the balls pass each other, they are at the same height at the same time t, so we equate the two:
    ## y = h - {1\over 2} gt^2##
    ## y = St - {1\over 2} gt^2##
    Obviously ##h = St## which gives us ##t##. We fill it in in the equation for the trajectory of one of the two and voila.

    Note that -- having seen the sketch -- we didn't make use of the symmetry, but we sure would do well to check that indeed ##t = \tau/2##.

    Re confidence in solving quadratic equations: well, this time you can get away with it, even though we are solving for the intersection between two parabolas. But it would be wise to pick up the necessary skills asap. Quadratic equations are the secret love of all exercise and exam authors, so they pop up all over the place in science education.

    Good luck !
     
  15. Feb 27, 2015 #14

    Suraj M

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    Mathias i think you might have complicated the method i suggested
    go at it like this:( if you want to)
    Let ##t_1## be the time taken by the ball dropped to reach a height x(measured from the ground) also x is the height at which the 2 balls meet.
    so you have $$x= ut₁ + ½gt²$$
    as u=0 you can get ##t_1=\sqrt{\frac{2x}{g}}##
    now consider the ball which was going up, let the time taken to reach a height ##x## be ##t_2##
    let ##t_{total}## be the time taken by the thrown ball to reach a height h where it's final velocity would become 0.
    so now ##t_2 = t_{total}-t_1## but you have to tell me why!
    so now you know t₁=t₂ so as BvU said and by these equations you can see that they would meet at ttotal/2
    I think you can do the rest.
     
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