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Final velocity for launched ball

  1. Jun 10, 2008 #1
    A ball (A) is launched straight down from a cliff while another (B) is launched straight up from the floor of the cliff. The final velocity for ball A is the inital velocity for ball b. Where will they cross paths? Up high, in the middle or down low?

    Also, 2 balls are launched upwards from the same spot but at different angles. the both will have the same max height. which ball goes further?
     
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  3. Jun 10, 2008 #2

    D H

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    What do you think? We don't do your homework here. Instead, we help you do your homework. That means we need to see something more than the question you were asked to solve.

    Show some work, please.
     
  4. Jun 10, 2008 #3
    what i think is happening is that on question 1, they are actually meeting in the middle since the intial velocity of the ball thrown from the floor is at the same final velocity of the ball dropped, do i make sense? the ball thrown from the floor is decreasing velocity while the ball dropped increases.

    A for the second one, i think that the ball with the greater angle will fall shorter than the ball with a less angle. the inital velocities are different i know that. Maybe i am thinking to much in the common sense part? i am completely confused about this one. I dont need answers i just need direction. As for both these questions, there are no numbers involved.
     
  5. Jun 10, 2008 #4
    2 questions(moved from previous posting)

    1. The problem statement, all variables and given/known data

    A ball (A) is launched straight down from a cliff while another (B) is launched straight up from the floor of the cliff. The final velocity for ball A is the inital velocity for ball b. Where will they cross paths? Up high, in the middle or down low?

    Also, 2 balls are launched upwards from the same spot but at different angles. the both will have the same max height. which ball goes further?


    2. Relevant equations



    3. The attempt at a solution

    what i think is happening is that on question 1, they are actually meeting in the middle since the intial velocity of the ball thrown from the floor is at the same final velocity of the ball dropped, do i make sense? the ball thrown from the floor is decreasing velocity while the ball dropped increases.

    A for the second one, i think that the ball with the greater angle will fall shorter than the ball with a less angle. the inital velocities are different i know that. Maybe i am thinking to much in the common sense part? i am completely confused about this one. I dont need answers i just need direction. As for both these questions, there are no numbers involved.
     
  6. Jun 10, 2008 #5

    Dick

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    Is this really a homework question, or are you just thinking about things? b) is pretty easy. Since they reach the same height they have the same initial component of y velocity and have the same time in the air. So the one with the greater initial component of x velocity will win. Which one is it? a) is (I think) much harder. I really can't think of a nice way to present it. But so far I seem to be finding that they will only meet in the middle if g=0.
     
  7. Jun 10, 2008 #6

    rl.bhat

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    When you say A ball (A) is launched straight down from a cliff , what is its initial velocity? Whether the ball A and B are launched simultaneously? In that case the meeting point will be decided by the initial velocity of B.
    In the second one, high speed low angle and low speed high angle can have same maximum height. The range will be more in the first case.
     
  8. Jun 10, 2008 #7

    Dick

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    "final velocity for ball A is the initial velocity for ball B". Call the initial velocity for A v0. Since they are initial velocities, yes, launched simultaneously. It's actually described pretty clearly. I think I know the answer, but I can't find a simple way to present it. Can you? The kinematical equations look creepy.
     
  9. Jun 10, 2008 #8

    D H

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    That does not mean you can't use variables to represent things like speeds, heights, angles. Doing physics without mathematics is a bit like playing a video game without an interface. It just doesn't make sense.

    How much time does it take for the dropped ball to fall half the distance? For the thrown ball to rise half the distance? Are these times the same, or different?

    The greater angle with respect to what? Vertical or horizontal?

    You might find it helpful to draw a picture for both problems.
     
  10. Jun 10, 2008 #9

    D H

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    Moderator: This thread should be merged with this one.

    Presumably there is gravity given that there are an initial and final velocity. Here is an easy way to look at it: They will meet in exactly half the time it takes for A to fall to the bottom of the cliff (or for B to rise to the top of the cliff). bigdaddy: prove this. If object A takes time t to fall a height h, how far will it have fallen during half of this time?
     
  11. Jun 11, 2008 #10

    Dick

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    Yes, should definitely be merged. But I think this is a 'hobby' question rather than a real homework question. I.e. question to debate over pitchers of beer. But it is a good question. I'm just guessing because of the obviousness of b) and the non-obviousness of a). If g=0, then they meet in the middle. This is easy and obvious and corresponds to the case v0 very large, g small. If v0=0, g>0 then they meet R fraction up the cliff. I'm not saying what R is just in case it is a real homework question. This corresponds to the case of g large and v0 small. This is easy but not completely obvious. In between must be in between R and 1/2. That's the best I've come up with.
     
  12. Jun 11, 2008 #11

    D H

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    Thanks, for the merge.

    (a) is just as obvious to me if you realize they must meet at the midpoint in time rather than the midpoint in space. The qualitative answer falls out immediately. This looks like a qualitative physics problem to me, so it might well be real.
     
  13. Jun 11, 2008 #12
    I found similar questions on the internet somewhere and with the questions it said there are more than one answer. I wanted to see if there really is more than one answer. Basically test opinions and see what i come up with. And really because of the lack of numbers, everyone really makes good points on either side of the equation. I think on question 1, if g>0 then they will meet lower to the ground say a 1/4 of the distance between the to launch spots, whereas if g<0 they will meet towards a higher point say 3/4. Am I right?

    I have to say on question 2, the ball that goes the farthest is the one with the greater initial x velocity. The angles are wrt the horizontal plane.
     
  14. Jun 11, 2008 #13

    Dick

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    That's exactly right. If you work out the extreme cases g~0 v0 large, you find the ratio is almost 1/2. If g is large and v0 small, the ratio is almost 3/4. So yes, upper half. I hadn't thought about g<0 case but what you say makes sense.
     
  15. Jun 11, 2008 #14

    D H

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    The balls only meet at 1/4 the distance down the cliff if s0 is zero. They always meet somewhere between [itex][1/4 h, 1/2 h)[/itex] if [itex]s_0 \ge 0[/itex] and [itex]g>0[/itex]. If [itex]g>0[/itex], the balls always meet above the midpoint regardless of whether the ball at the top of the cliff is is thrown up or down. I think we can eliminate the case of [itex]g \le 0[/itex] because of the way the problem is stated (i.e., words like cliff, down and up, bottom and top).

    The way to solve this problem qualitatively is to realize that the balls must meet at the midpoint in time. The trajectory of the ball thrown up from the bottom of the cliff is a time reversal of the trajectory of the ball thrown down from the top of the cliff: [itex]x_b(t) = x_a(t_f-t)[/itex] where [itex]x_a(t)[/itex] and [itex]x_b(t)[/itex] are positions of the two balls as functions of time and [itex]t_f[/itex] is the time the ball takes to fall down (or rise up) the full height of the cliff. At the midpoint in time ([itex]t = t_f/2[/itex]), [itex]x_b(t_f/2) = x_a(t_f-t_f/2) = x_a(t_f/2)[/itex]. Since the ball thrown from the top is accelerating downward, it will cover less distance down the cliff face for [itex]t\in[0,t_f/2][/itex] than it does for [itex]t\in[t_f/2,t_f][/itex]. So, they must meet somewhere in the upper half of the cliff.

    I can see people coming up with different answers for the second question because some will talk about angle from the vertical and others, angle from the horizontal, and do so without realizing they aren't talking about the same thing. The ambiguity goes away as soon as a reference line is established. The ambiguity also vanishes if the discussion is couched in terms of horizontal velocity component instead of angle.
     
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