1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding a object relative of time

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Person X jumps off a cliff from a height of h to the ground. On the way down X fires off a boaster which has a positive acceleration compared to gravity. Find in terms of t(time in seconds), mass, and gravity


    2. Relevant equations
    x=x0+(vt^2)+1/2(at^2)
    F=ma


    3. The attempt at a solution
    i chose this equation based on the fact that it uses time
    coord (x) will eventually = zero, or (x0) will = zero
    the initial velocity will also be zero, which brings me to the simplified equation of x = 1/2(at^2)
    With that in mind I can push everything around to sqrt(2a) = t
    but I haven't been able to solve for (a)
    Fnet=F(gravity)-F(propellent)

    a=(F(gravity)-F(propellent))/m
    a=(mg-F(propellent))/m

    I'm thinking i have to integrate the acceleration of the boaster due to it not being constant but I am just stuck.
     
  2. jcsd
  3. Sep 19, 2011 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have chosen the correct equation. However, you need to be careful about how you use it. If person X starts at [itex]x_0 = h[/itex] and finishes as [itex]x=0[/itex], then [itex]x-x_0 = -h[/itex], which is not zero as you have.

    The equation you have chosen is only valid for constant acceleration, so you can't use it if the acceleration isn't constant. However, what you can do is split the problem into two parts: (i) What happens before the person engages his booster and (ii) What happens after he engages his booster.
     
  4. Sep 19, 2011 #3
    I wasn't saying they were both zero but that one of them was zero. In which case it points at gravity's acceleration(being positive, or negative).

    Gravity is constant, where-as the boaster(propellent) is not constant. Before the boaster he falls at a constant acceleration, and the boaster would be a counter to that. Hence, the Fnet=F(gravity)-F(boaster).
    Since the boaster isn't a uniform for the equation I'm figuring that it has to be integrated but that is where I am stuck. The fact that the Force of the boaster isnt' uniform is why I didn't break it down in my OP.
     
  5. Sep 19, 2011 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I know you weren't saying that they were both zero. However, the equation you have written implies that they are both zero.
    Let's start again. Can you write down an equation for the height of the person above the ground before he engages his booster?
     
  6. Sep 19, 2011 #5
    h=1/2gt^2
    Or solving for time sqrt(2h/g)=t
     
  7. Sep 19, 2011 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No. At t=0, your height is zero, when it should be h.
     
  8. Sep 19, 2011 #7
    Then h would equal zero, and going in the same direction a g.
     
  9. Sep 19, 2011 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This makes no sense whatsoever. Let's go back to the beginning. The equation we are working with is

    [tex]x(t) = x_0 + v_0t + \frac{1}{2}at^2.[/tex]

    [itex]x(t)[/itex] is the position (height) of the person at time, time.
    [itex]x_0[/itex] is the initial position (height) of the person (i.e. [itex]x(0)[/itex]).
    [itex]v_0[/itex] is the initial velocity of the person.
    [itex]a[/itex] is the (constant) acceleration of the person.
    [itex]t[/itex] is time.

    Do you understand?

    If so, can you tell me in your case, when the person jumps off the cliff, but before he engages his booster, what each of those quantities are?
     
  10. Sep 19, 2011 #9
    At t=0 we can assume that we are starting at zero. Then at t=0 we would also be at zero with a velocity of zero, and zero acceleration. The only value we have is the acceleration of gravity, everything else is variable.
    This all about breaking down the equation
     
  11. Sep 19, 2011 #10

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can do that if you like, but it would make more sense to set [itex]x_0=h[/itex]. However, we'll stick with your way.
    Fine.
    No. The instant the person steps off the cliff, the acceleration is non-zero. What is the value of the acceleration?

    P.S. You didn't type the full question, what are we actually trying to find?
     
  12. Sep 19, 2011 #11
    I actually wrote it as it was in my text.
    The value of the acceleration would be gravity, or (g) until the booster engages.(note) this is why I was thinking of integration because the booster isn't constant but somehow figures in to this.
    P.S. We are not given any equations, or numbers to plug into this. Just the small word problem. what I am trying to figure out is the full equation
     
    Last edited: Sep 19, 2011
  13. Sep 19, 2011 #12

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you have copied the question exactly as given, then I'm afraid I don't know what to do as the question hasn't told up what to find.
     
  14. Sep 19, 2011 #13
    The question is find in "terms of time, mass, and gravity.""

    Ergo create an equation for the tiny story

    We're not looking for a specific height, time, mass, or acceleration but the big equation.

    Breaking it apart from what I can figure, which is the constant acceleration:

    t = sqrt(2h/g)

    the part I can't figure is the booster acceleration which isn't constant but has to figure in this equation
     
    Last edited: Sep 19, 2011
  15. Sep 19, 2011 #14
    my best guess as to the answer is: sqrt( (2h(mg-FB)/m) ) = t

    the problem is the units don't match

    I have the same issue with another problem:
    A helicopter is taking off vertically. the only forces acting upon it is gravity, and the air pressure from the blade(Fair) pushing it up. If the helicopter takes off at t=0 what is it's velocity at time t?

    it is basically the same problem with lots of unknowns. I still don't know how to find the Force portion of the equation.
     
  16. Sep 19, 2011 #15
    [Solved] Finding a object relative of time

    So in case someone else ever needs this problem the solution goes like this:


    sqrt((2hm)/(FB-mg)) = t

    if I had numbers to put into this then it would have been easier
     
  17. Sep 19, 2011 #16

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Then the problem was to find the time t in terms of the g, the mass, the height, and the booster force. But that is not what you wrote in the original post of the problem statement:

     
  18. Sep 19, 2011 #17
    True, but what i had posted was how the text book had it. And I had to figure out that the answer required those extra terms
     
    Last edited: Sep 19, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook