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Time is all you need

  1. Nov 1, 2007 #1
    Okay, so! My physics is lacking, but my life sciences are excellent, but that won't save me.

    How do I determine how long an object remains in the air and its velocity when it strikes the ground given:

    An object is thrown up into the air and then falls back to earth, remains in the air for 5 seconds before striking the ground; neglect air resistance.

    thanks in advance
     
  2. jcsd
  3. Nov 1, 2007 #2
    There is not really enough information in the question.

    From how high is the object thrown?

    If you knew that you could come to a solution.

    The object accelerates towards the ground at 9.8 m/s all the time (even when it is going up - this is what slows it down).

    The velocity when it strikes the ground is dependent upon how high it gets when it is thrown.
     
  4. Nov 1, 2007 #3
    that is all the information that was given.
     
  5. Nov 1, 2007 #4
    my teacher hinted at the fact that its original and final velocity will be 0 and that it takes 2.5 seconds to reach its maximum height and 2.5 more seconds to hit the ground. I think that is what she said.
     
  6. Nov 1, 2007 #5
    Does this make sense at all?

    displacement = initial velocity(change in time) + 1/2(acceleration)(change in time)^2
    so

    x = 0(10) + 1/2(9.81)(10)^2 ?
     
  7. Nov 1, 2007 #6
    OK, the velocities will not be zero, but will be equal.

    If she suggested that it takes 2.5s to reach max height and another 2.5 to hit ground, then it must be being thrown from ground level. (ie initial height =0)

    You should exploit the symmetry of the situation and split the problem into two halves: When the object is going up and when the object is coming down.

    Your equation is correct, but your substitution is wrong. Also, it may not be the right equation to use for this situation. What exactly are you trying to find?
    If it is the velocity as it strikes the ground then you most certainly need to use one of the other equations: you find it.

    Good luck.
     
  8. Nov 2, 2007 #7
    Hi. Considering the fact that the maximum height was not given, I do not think it is a good idea to exploit the symmetry of the situation. What do you think?
     
  9. Nov 2, 2007 #8
    The maximum height isn't really important to determine the simmetry of the movement, what is important to determine that is: if particle [tex]b[/tex] is shot from a height [tex]y_0[/tex], then if it lands at [tex]y[/tex], [tex]y_0 = y[/tex]. In this case, I guess we have to assume the particle is being thrown from [tex]y_0 = 0[/tex], and when it lands on Earth, it's also [tex]y = 0[/tex]. If we couldn't assume that, don't think the OP would even be able to solve the problem.
     
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