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Could you help me find what is wrong with these problems?

  1. Jul 22, 2007 #1
    1. The problem statement, all variables and given/known data
    1. A bullet is fired vertically upward and returns the ground in 20 seconds. Find the height it reaches.


    2. Relevant equations
    1. d=g(t^2)


    3. The attempt at a solution
    1. =9.8m/s^2(20s^2)
    =9.8m/s^2(400s)
    d=3,920m upward
     
  2. jcsd
  3. Jul 22, 2007 #2

    cristo

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    The questions says the bullet returns to the ground in 20s, but you use 20s to find the total distance. How long does it take to reach its maximum height? Knowing this, can you find the height?
     
  4. Jul 22, 2007 #3
    Actually, that is where I am having a hard time, I just don't know what equation to use to further find the answer... I hope someone can help me... please
     
  5. Jul 22, 2007 #4

    cristo

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    Imagine you were plotting a graph of the displacement of the ball against time taken. What shape would this graph be? Does this help you spot any symmetry from which you can divide the total time into a time taken on the way up, and a time taken on the way down?

    (This is all presuming that the bullet is shot from a gun on the ground, and not raised by a certain height.)
     
  6. Jul 22, 2007 #5
    And what are d and t in the equation you used to calculate d? How was that equation derived and what meaning do d and t have? Think about this a bit and then apply the equation again. (Do this after you've done what cristo said)
     
  7. Jul 22, 2007 #6
    please cold you just give me the answer that you were ale to derive...coz i'm burn out from all the projects i have to pass this week...
     
  8. Jul 22, 2007 #7
    Yes, we could, but it would keep you from understanding the situation. What are your thoughts on this?

    A ball is thrown up and it takes 20 seconds to come down. Obviously, it rises a distance, stops (why?) and then drops down to earth. How much time does it take to rise up through that distance before stopping, and how much time does it take to descend from the point where it stops momentarily?

    EDIT: I got your pm. Well, okay we'll throw a huge hint...I don't want to give it away but I think I might:

    At the topmost point, the bullet stops momentarily, so its speed = 0. Suppose the initial speed = u and the time taken to rise to this point is [itex]t_1[/itex]. Can you relate u and [itex]t_1[/itex]? Can you find an expression for the height of this topmost point H?

    Now, for the descent, the bullet starts out with zero speed and has to cover a distance H in the reverse direction. Can you relate the time taken to come down, [itex]t_{2}[/itex] to the height H? Do you now see any symmetry in the problem?
     
    Last edited: Jul 22, 2007
  9. Jul 22, 2007 #8
    coz of gravity pull...irregardless of air resistance=you get a symmetrical diagram...it will take 10s to rise up and another 10s to go down...?
     
  10. Jul 22, 2007 #9
    Precisely!

    So you can do it in a few ways. One way is to compute the initial speed u, using

    [tex]v = u + at[/tex]

    with v = 0, u = u, a = -g, t = 10s. This gives u. Now, using

    [tex]v^2 = u^2 + 2as[/tex]

    with v = 0, u = obtained from the first step, a = -g, s = H. This gives

    [tex]H = \frac{u^2}{2g}[/tex]

    Alternatively, using

    [tex]s = ut + \frac{1}{2}at^2[/tex]

    for the upward rise, a = -g, s = H, gives

    [tex]H = ut - (1/2)gt^{2}[/tex]

    Get H from this.
     
  11. Jul 22, 2007 #10


    thnx...:smile:
     
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