1. Not finding help here? Sign up for a free 30min 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!

Free Fall With Air Resistance Problem

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = -g + bv. After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant b? (b) What is the acceleration at t = 0? (c) What is the accleration when the speed is 0.150 m/s?

    2. Relevant equations

    ƩF=ma
    W = weight
    FD = resistant force

    3. The attempt at a solution
    I drew a free body diagram with the resistance force pointing up and the weight pointing down, where up is positive.

    ƩF=W-FD=ma
    -mg + bv = ma
    -(1)g + bv = (1)a
    -g + bv = a
    bv-g = 0
    b = g/v

    I am stuck here because I am not sure how to find the terminal velocity. Do I do v=x/t=0.05/0.150=10/3≈3.33 m/s??? That would make b = 9.8/(10/3)=2.94Kg/s, which seems too high...Can anyone help? Thanks.
     
  2. jcsd
  3. Apr 29, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What distance does the Styrofarm travel with the terminal speed?

    ehild
     
  4. Apr 29, 2012 #3
    The terminal speed is when the acceleration is zero.
     
  5. Jul 1, 2012 #4
    During the first 0.5 meter fall, the velocity increases from 0 m/s to the terminal velocity. After the object falls 0.5 meter, the velocity remains constant.

    The velocity of the object as it falls the last 1.5 meters is the terminal velocity.
    Distance = terminal velocity * time
    1.5 = terminal velocity * 5
    Terminal velocity = 1.5 ÷ 5 = 0.3 m/s

    This is the velocity of the object, after falling the first 0.5 meter.

    a = -9.8 + bv
    At t = 0, v = 0
    a = -9.8

    At terminal velocity, acceleration= 0
    -9.8 + bv = 0
    -9.8 = b * v
    b = -9.8 ÷ v
    v = terminal velocity
    b = 9.8 ÷ 0.3
    b = 32⅔ = 32.667

    Now that we know the value of b, we can write the equation for acceleration.

    a = -9.8 + 32.667 * v

    (c) What is the accleration when the speed is 0.150 m/s?
    Terminal velocity = 0.3 m/s
    So this speed occurs during the time, when the object was accelerating/
    Use the equation for acceleration!
    a = -9.8 + 32.667 * 0.150
    a = -4.9 m/s^2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook