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Formula to find height required to reach x speed in freefall

  1. Jul 26, 2012 #1
    Acceleration (a or 9.8 for simplicity); initial velocity (vf); distance (d); Initial velocity (vi)

    trying to find a formula to answer "how far would you have to fall to reach x speed" where x is input from user.

    Is it possible to answer this using two steps, first find time then find distance?

    d = vi*t + 0.5*a*(t^2) // t isn't given (can it be found with vi,vf,a ?)
    d = (0) + .5(9.8)(t^2)
    d =4.9(t^2)

    vf = a(t) + vi // seems it can
    vf = 9.8(t) + 0
    t= (vf-vi) / a
    [tex]t = \frac{vf-vi}{a} [/tex]

    yea that worked; code ended up being

    //first we must find time
    time = speed / 9.806;
    //now use time in a distance equation
    height = (vi*time) + (0.5*9.8)*(time*time);


    //if you jump out of a plane going towards ground at 250mph , how long/far would it take to reach 120mph (or terminal velocity in a traditional skydiving formation) due to air resistance , what force is being applied to your body?

    That's the next part I need help with.
    Last edited: Jul 26, 2012
  2. jcsd
  3. Jul 26, 2012 #2

    Simon Bridge

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    Welcome to PF;
    Well done - just some pointers:

    The way to be sure of yourself in these things is to start by drawing the velocity-time graph of the motion.

    You are better to do the calculation in one step - avoids rounding errors in between stages. This is a good discipline for later.

    Note: good to see you using latex - in latex you can use an underscore to make a subscript so your equation markup would be T=\frac{v_f - v_i}{a} to give you[tex]T=\frac{v_f - v_i}{a}[/tex]... you can substitute that directly into the equation for d. (Note: I've used T instead of t to avoid confusion with the label for the time axis.)

    the v-t graph is a triangle if the initial speed is zero - the distance is the area of the triangle (base=T, height = vf right?) so [tex]d=\frac{v_f T}{2}=\frac{v_f^2}{2a}[/tex]

    (of course you could have just remembered the kinematic equation: [itex]v_f^2 = v_i^2 + 2ad[/itex]
  4. Jul 26, 2012 #3
    after experimenting a little I can conclude I went wrong somewhere.
    If I set initial velocity to 0 calculations work, if initial velocity !=0 , answer is incorrect.

    Looking at the formula it's easy to see if vi > 0 the height increases - opposite of what should happen. I'm thinking I need to input something as a negative - however in my head vi / vf and acceleration are traveling in the same direction, meaning if one was negative they all would be.

    I vaguely remember something about velocity not having a direction


    the other equation
    to isolate d
    [tex] d = \frac{v_f^2 - v_i^2}{2a} [/tex]
    wanting to double check that's right

    order of operations becomes a problem for the compiler the bigger the equations get but I can see the value in having a single step equation.
  5. Jul 27, 2012 #4
    for the red...

    You have to establish a direction to use these kinematic equations. Velocity does have direction. As for doing this on a computer with a language or whatever, you have to use signs somehow.

    The second part of your question air friction acts up, gravity down so you have two forces acting on you. The problem is that air resistance is velocity dependent. In other words, as you fall the acceleration is going to change. And you cannot use the equations above with changing acceleration unless you modify. The force of air is usually modeled to be kv^2 or kv. So basically you get something like Kv^2 - mg = ma but a is changing... so the distance you fall cannot be done with the kinematic equations in which accel. does not change, its tougher.
  6. Jul 27, 2012 #5
    In this question it is subtly implied that acceleration is not constant. Terminal velocity is attained after falling for some distance and reaching an equilibrium velocity when your forces (drag and gravity) are equal. If you've had fluids than you can assume the air is inviscid and incompressible and you may solve this equation for z (the distance you fell):

    [itex]P + 1/2 \rho V^2 +\rho g z = 0[/itex]
    P = thermodynamic pressure, dependent on temperature and density primarily
    ρ = density of the air
    z = distance of freefall (really z-z0)

    Unless you've had fluid mechanics or an intro to fluids class I doubt that your teacher will expect you to use this equation. Should you decide to go the extra mile you can find its reference page on Wikipedia as the "Bernoulli equation".

    This is for the velocity of a fluid (specifically a streamline). To find the velocity of a falling body you can also use the terminal velocity formula, which is found on Wikipedia and balance it with gravity. Unfortunately, this wont do you any good in finding the distance fallen.
    Last edited: Jul 27, 2012
  7. Jul 27, 2012 #6

    Simon Bridge

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    I think it could be read either way - context is everything. In the context that the only input parameter is a final speed, it seems reasonable to guess free-fall without air friction. After all, if you fell from orbit, you will go through a deceleration phase as well (possible you accumulated speed from high altitude will be faster than the terminal speed for lower altitude).

    In the event that initial velocity is not zero ... (setting the positive direction to "down") then there are several possibilities for the answer (and depending on how you interpret "fallen"). If the initial velocity is non-zero, and upwards, then it is negative. But the input parameter is given as "speed" so the object could reach that speed on the way up, as well as on the way down. The distance "fallen" could be the total distance travelled in the trip or just the relative displacement (from the initial position).

    OTOH: the initial velocity is not one of the stated input parameters ... it seems reasonable to consider it to be about falling from rest. I have the same reservation about air resistance, since altitude is not a parameter, the model would have to be simple anyway. For that matter, altitude is important for the value of g. Perhaps they just want motion with constant acceleration as a simply dynamical system?

    Since this is a computer programming exercise - more care should be taken about how we go about it. What is the education purpose of the exercise? What do they want you to learn? It could be you are supposed to use an iterative or recursive approach rather than just bunging the number into the kinematic equations. This is why context is so important.
  8. Jul 27, 2012 #7
    /if you jump out of a plane going towards ground at 250mph , how long/far would it take to reach 120mph (or terminal velocity in a traditional skydiving formation) due to air resistance , what force is being applied to your body?

    Yes you have to be right on this one Simon. They would not put something straightforward and then hit him/her with a right cross. Although I have run into that, but seeing what the intended purpose is you have got to be right. Constant force for air so constant acceleration.
  9. Jul 27, 2012 #8
    If this is a computational problem (as mentioned above) then summing the forces and solving for the velocity numerically is a straight forward process, for a cs major at least.
  10. Jul 27, 2012 #9
    Here is a huge hint:
    If you sum your forces in the y direction you will get

    [itex]1/2 \rho (\Delta v)^2 S C_D - m ({\Delta v}/{\Delta t}) = mg[/itex]
    [itex]1/2 \rho (\Delta v)^2 S C_D \Delta t - m ({\Delta v}) = mg \Delta t[/itex]

    You can solve for [itex]\Delta v [/itex] numerically using the quadratic equation.

    To obtain the unknowns S and CD, use the terminal velocity equation to solve for their product S*CD

    Once you have all of the velocities for each iteration simply perform a numerical integration to get your distance.
  11. Jul 28, 2012 #10

    Simon Bridge

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    Cool - remains only for OP to confirm :)
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