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Another Kinematics Problem (I have solution)

  1. Aug 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A parachutist bails out and freely falls 59 m. Then the parachute opens, and thereafter she decelerates at 2.7 m/s2. She reaches the ground with a speed of 3.1 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?

    2. Relevant equations
    q8GGHtZ.png



    3. The attempt at a solution
    59=0+1/2(9.8)t^2
    t=3.47
    Vf=Vo +gt
    Vf=0+9.8(3.47)
    Vf=34.01
    _________
    Vf^2=Vo^2+2(-2.7)(Y)
    3.1^2=34.01^2-5.4(Y)
    Y=212.42

    212.42=34.01(t)-1.35(t)^2
    I got 2 answers 13.7445 and 11.448 (due to quadratic formula)
    they both equate to 212.42 so I need to choose and why. I got other answer correct except the t
     
  2. jcsd
  3. Aug 9, 2015 #2

    SteamKing

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    You need to show units in your calculations. It makes it easier for others to check your work. You might also save having points deducted on exams for not showing units.

    That said, why did you solve a quadratic to find the time elapsed after the parachute opens?

    You are given the deceleration and the final velocity on striking the ground. You calculated the velocity at the end of freefall. This information is tailor-made for substituting into your first kinematics equation to find the time.
     
  4. Aug 9, 2015 #3
    Y
    Hey that's right sir I have bad habits not posting units in exams (which lead to deduction lol)

    I tried the other method and I got 13.7445s and I add it to 3.47s = 17.21s but it is wrong and can you please explain where did I get wrong but my (b) is right (271.42m)

    EDIT: I kinda have bad english so sorry
     
  5. Aug 9, 2015 #4

    SteamKing

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    According to the OP, the parachutist decelerates at 2.7 m/s2 and lands with a velocity of 3.1 m/s.

    Your calculations show that at the end of freefall, the parachutist has a velocity of 34.01 m/s.

    I don't know how you solved the equation and got a time of 13.74 s to do this. Please show your work.
     
  6. Aug 9, 2015 #5

    212.42=34.01(t)-1.35(t)^2
    I just use quadraic 13.744 and 11.45

    I don't know what you are saying but I shown my work already in the solution area
    I completed the template.
     
  7. Aug 9, 2015 #6
    I got 212.42=34.01(t) -1.35(t)^2 from the 3rd formula in my OP
     
  8. Aug 9, 2015 #7

    SteamKing

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    My original point was that solving the quadratic equation was unnecessary.

    The first kinematic equation, v = v0 + at, works perfectly well for finding how long it takes for the parachutist to decelerate from freefall to a speed of 3.1 m/s on landing.
    You know everything you need to find the time, t, as explained in Post #4.
     
  9. Aug 9, 2015 #8
    Yeap I have no problem answering the problem until the quadratic.. So I tried using a calculator and the answer shows the 11.448.
     
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