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Newtons Second Law Formula Question

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    I've been given a problem that an object falls from a distance of 800 feet with an initial velocity of 4 f/s. Gravity acts on the object, but air resistance can be disregarded. Find the location at time t and find when the object hits the ground.


    3. The attempt at a solution

    I am confused on the formula. Is the formula that I need to use: [tex]\frac{dv}{dt} = g - \frac{bv}{m}[/tex] ?

    So with all the numbers plugged in the formula would be:

    [tex]\frac{dv}{dt} = 9.8 - \frac{12}{m}[/tex] ?


    Then I can integrate this and solve to get the answer?

    [tex]\frac{dv}{dt} + \frac{12}{m} = 9.8[/tex]

    Mass is not given, so what do I do about the m? I'm completely lost and I just need a hint or two about how to setup the formula.

    Thanks.
     
  2. jcsd
  3. Oct 10, 2009 #2

    Mark44

    Staff: Mentor

    Why would you use this formula? [tex]\frac{dv}{dt} = g - \frac{bv}{m}[/tex]
    This says that a = g - something. Since you are told you can ignore air resistance, what could possibly be acting on the object to reduce its acceleration?

    Isn't dv/dt simply going to be g?
     
  4. Oct 10, 2009 #3
    Hmm... it seems it would be, but how would I set-up the formula to determine where the object is at time t? I can't do that with just dv/dt = 9.8, can I?

    The at time = 0, velocity is 4, so assuming no air resistance, does that mean it takes 125 seconds to reach the bottom, or does gravity alter its speed?
     
  5. Oct 10, 2009 #4

    Mark44

    Staff: Mentor

    You have dv/dt = -9.8. (I'm choosing the upward direction to be positive.) How can you solve this differential equation to get v as a function of t? Don't forget that v(0) = -4 ft/sec. You didn't say, but I'm assuming you meant that the initial velocity was downward.

    After you get v(t), you can replace v(t) by ds/dt, and do something with that to get the position s as a function of t. This time, don't forget that s(0) = + 800 ft.
     
  6. Oct 10, 2009 #5
    Hmm.. so to find v, I would integrate dv/dt, which would give me: v = 9.8t + C? Then, to solve for c, I do 4 = c and that gives me the equation for time which is v(t) = 9.8t + 4?

    Is this correct so far, or what do I need to reconsider?
     
  7. Oct 11, 2009 #6

    Mark44

    Staff: Mentor

    No, not correct, but you have the basic idea. You need to pay attention to the signs of things, though. The way I have set it up, down is negative and up is positive.
     
  8. Oct 11, 2009 #7
    Okay, so would it just be v(t)= -9.8t + 4 and then I substitute v(t) with ds/dt?
     
  9. Oct 11, 2009 #8

    Mark44

    Staff: Mentor

    No, it's partly wrong. Which direction is the initial velocity? Reread post #4 carefully.
     
  10. Oct 11, 2009 #9
    Oh, yeah, whoops. It is: v(t)= -9.8t - 4. Okay, now I make the problem:

    ds/dt = -9.8t - 4

    Am I on the right track now?

    I am suppose to integrate next, correct?
     
  11. Oct 11, 2009 #10

    Mark44

    Staff: Mentor

    Yes and yes.
     
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