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Homework Help: Vertical motion Kinematics Confused as heck

  1. Aug 27, 2011 #1
    Frederick comes up with the following scheme to measure a building's height. A timing mechanism is set up which measures the time taken by an object dropped from the top of the building to fall the last 2.0 m before it hits the ground. The object is found to take 0.125 s to move the last 2.0 m. How high is the building?
  2. jcsd
  3. Aug 27, 2011 #2
    There are three important points on the track of the ball:
    When it's at the top of the building,
    when it's at the height of 2 meters above the ground,
    and when it arrives the ground.

    Try to calculate the ball's speed exactly when he is two meters above the ground. There's only one appropriate motion equation you could use here.

    Then use this speed to determine the buildings height. Assume that the ball had the initial velocity of 0 on the top of the building (I guess that's what they mean).

    Good luck!
  4. Aug 27, 2011 #3
    I tried this but it was incorrect :/
    v=s/t=2/0.125= 16m/s, then substituted into
    v^2=u^2 +2as, rearranged to make the displacement by itself, so..
    s=(v^2-u^2)/2a=16^2/2*-9.81= -13m
    I looked at the answer and i was wrong so i tried another method...
    i assumed the velocity 2m above the ground was 16m/s (velocity i got from original equation).
    so i said, v=u + at=16 + 9.82*0.125= 17.22m/s then substituted into the equation...
    s= v^2=u^2+2as rearranged for displacement so..
    I checked the answer again and it was wrong.
    Im not sure how to do it but the answer is 14.1m
  5. Aug 27, 2011 #4
    whoa whoa whoa :)

    s = v*t doesn't suit this problem at all, since there's gravity involved and the ball is therefore accelerated. This formula is only valid when there are no forces in action.

    Wanna show me a correct attempt with a valid formula?
  6. Aug 27, 2011 #5
    i honestly dont know could you help me?
  7. Aug 27, 2011 #6
    For a constant acceleration (like gravity for small heights - free fall), this formula applies:

    x(t) = xo + v0t + 0.5at2

    Where x0 is the initial position, v0 the initial velocity and a the acceleration.

    Do what you did, you did it right - only use this formula instead. I hope it isn't the first time you see it :-)

    Make sure you understand that the term "initial velocity" is always in reference to a certain segment you define. For example the initial velocity could be 0 if your starting point is at the top of the building, but could be something else if your starting point is somewhere else...

    Can you make it now?
  8. Aug 27, 2011 #7
    umm possibly what does x(t) mean?
  9. Aug 27, 2011 #8
    Actually i dont think i can figure this out without you:/
  10. Aug 27, 2011 #9
    heh, I'm your only hope? :-)
    I would first of all want to recommend you study this properly and review this material. It's not complicated and is very important.
    Do you know the formula I've written? That's what you use in the case of accelerated (constantly) bodies. Free falling bodies is a classic example.

    I won't do the calculations for you but I'll guide you.

    1. x(t) is the position of the object at the time t. you could just call it x. :)
    You know that the ball traveled 2 meters in 0.125 sec. You know that the acceleration is a = 9.8 m/s^2(we define the direction from up to bottom to be the positive axis direction for simplicity).
    So the ball starts at "0", and travels 2 meters. Its final position is therefore x(t) = 2. Plug in everything you know and get the "initial velocity" (that's the velocity when it's two meters above the ground) from the formula.

    2. Take it further exactly like you showed me.

    Make sure you understand why a = 9.8 and not -9.8: It is because of the way we defined that "local axis" to start two meters above the ground and to be pointing downwards.
    Of course you could also choose it differently: You could say that the axis starts on the ground, and that it's pointing upwards.
    Then x(t) = 0, but x_0 (initial position) would be 2, whereas a = -9.8, because it's pointing against the direction of the axis.

    Hope it's clearer!
  11. Aug 27, 2011 #10
    holy moly im lost :/ can you do the equation? i dont know how to do it, I just can't understand it :/ if you show me it will make it clear

    Thanks btw especially for putting up with my lack of knowledge.
  12. Aug 27, 2011 #11


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    Just checking what you can calculate.

    from what height must you drop an object in order that it reaches a speed of 22 m/s.
  13. Aug 27, 2011 #12


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    As Tomer has said, all you need is the equation: x(t) = x0+v0t+0.5at2

    You just need to apply it to your problem. Do it in two steps if that makes it easier. First off, the object starts from rest at the top of the building and accelerates downward until it gets to 2m above the ground. From there is has some velocity which it has gained from falling, and it then moves the last 2m, still being accelerated by gravity.

    The time the object takes to fall the last 2m gives you the velocity it had when it was at 2m. Then you can use this to get the height of the building.
  14. Aug 27, 2011 #13
    I believe you when you say you're lost, but I think it's against the forum's rules to actually solve it for you.

    I've really written exactly what it is you need to do in order to get that velocity you seek. You need to understand the problem, it's pretty simple. Draw it to yourself.
    Show us an attempt. I would be then much easier helping you. I'm sorry for being a pain.
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