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Finding Distance

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    This is a little confusing for me but im pretty sure I need to look at it in a different way so Im asking for some advice here is the situation....


    Basically I have to find the height so the golf ball goes down the ramp and lands on the ground exactly 1meter away from the table.

    2. Relevant equations

    For equations I really don't know which one to use I was thinking about using

    Vf² = Vi² + 2a (displacement)

    3. The attempt at a solution

    I know the height from the table to the ground is 92cm
    The table top is 204cm longs
    When the golf ball is released it rolls on the table top for 2cm and then falls to the ground
    The place where the golf ball lands is 1meter away
    The golf ball is 45g
    the system is frictionless
    I think gravity plays a part on how to solve it so 9.8 = g

    I still didn't use the equation yet because I'm not really sure if its the right now, I was probably thinking I might need to use trig functions because It resemebles a triangle, but if i not I still have no clue on which one.

    Any help on this would be highly appreciated, I don't want the answer because to me its more important I know the process on how to get the answer, rather I would just like an equation to use and little hints on things.

    ***Sorry about the wrong title, its suppose to be "Finding Height", and if it helps we are currently learning about momentum

    Last edited: Feb 27, 2009
  2. jcsd
  3. Feb 27, 2009 #2


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    First figure how long it takes a ball to drop .92m.

    If you know how long it takes to drop, you know how much horizontal velocity it needs to get there.

    Once you know how fast it needs to be, then figure how much acceleration it needs down the ramp.

    As it turns out the acceleration you need is the same percentage of gravity, that the height x is to the length of the ramp. (You should satisfy yourself on that point.)
  4. Feb 27, 2009 #3


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    By using the conservation of energy, you can find the velocity with which the golf ball reaches the table top.
    What will be the direction of the velocity of the golf ball?
    When the golf ball leaves the table, what will be its direction?
    How much time it takes to reach the ground?
  5. Feb 28, 2009 #4
    Ah sorry it took a while for me to reply, and thanks for the tips so far I got this far

    Vf² = Vi² + 2a (displacement)
    Vf² = 0 + 2 (10 or 9.8) (0.92)
    Vf² = 20 (0.92)
    Vf² = 18.4
    Vf = 4.3m/s

    Now I am thinking I am suppose to find time so I plan to use this equation:

    displacement = Vi (t) + 1/2 a t²

    0.92 = 0 + 1/2 (10) t²
    0.92 = 5 t²
    0.2 = t²
    0.45 s = t

    OR i was maybe thinking since I have the velocity now I can find momentum and use this equation :

    delta P = Force delta t

    Which would be like this when I worked it out

    p =mv
    p = 0.045kg (4.3 -0 )
    p = 0.2 kg m/s

    delta p = F delta t
    0.2 = 0.045kg (10 or 9.8) t
    0.2 = 0.45 t
    t = 0.4s

    if I am on the right path so far let me know, but to be honest I am really getting confused and would like a little more help

    Last edited: Feb 28, 2009
  6. Feb 28, 2009 #5


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    No. Your first step is not the Yellow Brick Road to the Solution.

    Your second step however should have yielded you useful insight.

    Rather than churning out all the equations you know, think about what you need.

    In your second step you apparently found the time that it takes for the object to fall. By your drawing note that the velocity at the bottom of the ramp gets straightened out to be all horizontal, or at least I presume that there is enough flat table to straighten it out. If your horizontal velocity then is determined by the speed at the bottom of the ramp, then you can figure out how fast it needs to be to go 1 m.

    You have .45s (Using 9.8 I get .433s) to go 1 meter. How fast does it need to be then?

    Knowing how fast it needs to be then you can figure what acceleration gives that to you over the distance of the ramp.
  7. Feb 28, 2009 #6
    Ah ok I think I know what you are trying to say

    You want me to find the acceleration then? If so here it is

    a = delta v / delta t
    a = 4.3 - 0 / 0.45
    a = 4.3 / 0.45
    a = 9.5 m/s ²

    If not could explain it in simpler terms or tell me what I need to actually do

  8. Feb 28, 2009 #7


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    Not exactly.

    The ball has to land where? 1 m away from the table?

    How long have you calculated it to drop from the top of the table to the floor? .45s looks to me.

    What velocity gives you 1m covered in .45 sec?

    You must have that answer before you can talk about any acceleration - i.e. how you got the ball to that velocity.
  9. Feb 28, 2009 #8
    What do you mean what velocity gives me 1m covered in 0.45s? Is it final velocity? If so then i just have to plug it in and solve for Vf?
  10. Feb 28, 2009 #9


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    Horizontally it has to travel 1 m in the .45s it has to fall. Is that not correct?

    When it comes down the ramp it will have a velocity that when it runs over the 2cm lavel part of the table makes the velocity all horizontal doesn't it?
  11. Feb 28, 2009 #10
    well the thing is, when it golf ball goes down the ramp and travels 2cm across the table and then falls, it has to land exactly 1m away from the table.

    So what do I have to find then now? Do i just use all the information I have and find the x ( the height for the ramp) ? Or is there something else I have to find
  12. Feb 28, 2009 #11


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    Until you answer what the horizontal velocity is off the table, you won't be able to solve the problem, because that condition must be met for it to hit 1 m away.
  13. Feb 28, 2009 #12
    So im guessing the horizontal velocity is the same as Vf

    vf² = vi² + 2a displacement
    vf ²= 0 + 2 (10) (1.02)
    vf ²= 20 (1.02)
    vf² = 20.4
    vf = 4.5 m/s

    Is that it? Or is it something else

    reason I am using 1.02m is because I added the 1m + 2cm the golf ball travels on the table if I just use the 1m instead the answer for vf = 4.47 = 4.5
  14. Feb 28, 2009 #13


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    No. No. and No.

    How far will the ball travel horizontally at 4.5 m/s after .45 sec? It will land 2 m from the bottom of the table is how far. So forget trying to drag that result back into your thinking at every turn because 4.5m/s gives the wrong result guaranteed.

    You need to answer the question about how fast it must be going to only go 1 m in .45 sec before you can worry about how it got to going that fast.
  15. Mar 1, 2009 #14
    how fast it must be going I guess you mean the speed?

    So its

    s = d/t
    s = 1 m / 0.45
    s = 2.2 m/s

    Is that the answer to your question now?
  16. Mar 1, 2009 #15


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    That means then that at the bottom of your ramp it must be going at 2.2 m/s. Now you can figure the acceleration down the ramp starting from rest.
  17. Mar 1, 2009 #16
    So its doing down the ramp I just the distance of the ramp?

    displacement = Vi t + 1/2 a t²
    2.15 m = 0 + 0.5 a 2.2 ²
    2.15 m = 0.5 *4.84 a
    2.15 m = 2.42 a
    a = 0.9 m/s
  18. Mar 1, 2009 #17


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    And where did the time come from?

    Don't you want to use this equation that were so eager to use before?
    Vf² = Vi² + 2a (displacement)
  19. Mar 1, 2009 #18
    Ah man, i was thinking about using it but every time I did you said it was wrong haha,

    vf² = Vi² + 2a (2.15m)
    4.3 = 0 + 4.3 a
    18.5 = 4.3 a
    a = 4.3 m/s²

    But i think the vf i used is wrong i think it should be something like this

    vf² = Vi² + 2a (displacement)
    vf² = 0 + 2 (9.8) (2.15m)
    vf² = 42.14
    vf = 6.5 m/s

    vf² = Vi² + 2a (2.15m)
    6.5² = 0 + 2 a (2.15 m)
    42.3 = 4.3 a
    a = 9.8 m/s²
    Last edited: Mar 1, 2009
  20. Mar 1, 2009 #19


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    No. You found Vf earlier.

    It is 2.2 m/s.

    You keep discarding what you've already found.
  21. Mar 1, 2009 #20
    Oh man, this is really making me feel slow sorry for the trouble

    vf² = Vi² + 2a (2.15m)
    2.2² = 0 + 2 a (2.15 m)
    4.84 = 4.3 a
    a = 1.13 m/s²
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