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Catching a glass falling to the ground

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data

    A person trips against a table, causing a glass to fall off the edge. An excellent human reaction time is 0.25 seconds. In that time, how far wil the glass fall? Assume that immediately (after 0.25 s has elapsed) begin to acclerate your hand so that you grab the glass when it is 6 in from the floor. What constant acceleration was necessary, and how fast was your hand traveling when you contacted the glass? Assume your hand moved in a straight line.
    96DN6.png
    2. Relevant equations



    3. The attempt at a solution

    I find the position vector of the hand to be:
    81ZwS.jpg
    I get the position vector of the glass to be:
    VZQmD.jpg
    I need to find the time it takes for the glass to fall to y=0.5:
    Ybggo.jpg
    So the hand only has 0.46 seconds to reach the glass. However, when I plug this into the position vector of the hand to find a_x and a_y and find the magnitude of the acceleration, I get like twenty-something ft/s^2 which is far below the correct answer of 241 ft/s^2. What am I doing wrong?!
    UtBcm.jpg

    Please help asap!
     
  2. jcsd
  3. Sep 8, 2011 #2

    PhanthomJay

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    Gold Member

    Good afternoon, Mr. President.

    Your equations, sir, with all due respect, made me think I was looking at Wiki.

    Keep it simple. Find the time it takes for the glass to be 6 inches from the floor, using distance = 1/2gt^2. Then that time less the reaction time is the time that your hand has to reach the glass, travelling a diagonal distance as found from Pythagorus' theorem.

    Warmest regards,
    Jay
     
  4. Sep 8, 2011 #3
    I agree, keep it simple.

    1) How far will the glass fall in 0.25 seconds?

    You know the value of gravity (9.81 m/s^2), and you know the time (0.25 seconds). Use the equation:

    S[itex]_{y}[/itex] = ut + [itex]\frac{at^2}{2}[/itex]

    The second part has already been answered.
     
  5. Sep 8, 2011 #4
    But that's what I did...
    y=(1/2)gt^2
    t=sqrt(2y/g)
    =sqrt((2)(2.5)/9.8)
    = 0.71

    0.71-0.25 = .46...

    Diagonal distance=2.5,
    y=(1/2)at^2
    a=(2y)/(t^2)
    =(2*2.5)/(0.46^2)
    = 23.63ft/s^2...which is what I got before.
     
  6. Sep 8, 2011 #5

    PhanthomJay

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    Oh, sure, that's a lot better than all those i's and j's.. But in the USA system, g is not established in units of m/sec^2, but rather, in units of ft/sec^2.


    9.8m/sec^2 = appx. 32.2 ft.sec^2.:frown:
     
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