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Skydiver problem

  1. May 9, 2004 #1
    Ok here's the thing.... i have an assignment due soon... like 3 days, and ive been sitting here trying to figure out this mutha of a question... i figured out one bit of it and yeah still not enough... anyways hope you people can help us =)


    QUESTION

    A skydiver weighing 70kg jumps from an aeroplane at an altitude of 700 metres and falls for T1 seconds before pulling the ripcord of the parachute.

    A landing is said to be gentle if the velocity on impact is no more than the impact velocity of an object falling at a height of 6 meters.

    The distance the skydiver falls during t seconds can be found from Newton's Second Law, F=ma.

    During the freefall portion of the jump, we will assume that there is no air resistance, so F=-mg where g=9.8m/s^2 and m=70kg.

    After the parachute opens, a signifigant drag term due to air resistance of the parachute affects force F, causing it to be F=-mg-kv where v is the velocity and k=110kg/seconds is a drag coefficient.

    (A) Find the range of times T1 at which the ripcord can be pulled for a gentle landing
    (B) Find the height after T1 seconds of free fall....


    ok i figured out that the gentle landing velocity has to be less than 10.8m/s and it takes 1.1seconds for it to fall 6m...

    can someone help by pointing me in the right direction into finding the velocity during the drag and the displacement of the parachuter when t = T1 ???

    thanks in advance! oh yeah sorry about the other post in the sub-forums... too bad i cant delete my own posts...
     
  2. jcsd
  3. May 9, 2004 #2

    Gokul43201

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    Have you heard the term "terminal velocity" ? What does it refer to ? Is it relevant here ?

    What happens to the diver's velocity, if when he pulls the ripcord ,
    a) He is falling faster than the terminal velocity,
    b) he is slower

    Compare your calculated terminal velocity with your calculated 'gentle landing' velocity.
    Do either of the above conditions guarantee a gentle landing? What about the other case ?

    I believe there's enough here to get you going.
     
  4. May 9, 2004 #3

    HallsofIvy

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    Just do it step by step. During the free fall, F= ma F= mg so a= g (of course). Falling at a constant acceleration of g= 9.8 m/s2, his speed after T1 seconds is -9.8T1 m/s and he will have fallen -4.9T12 meters.

    Now the rest of the problem is basically solving a differential equation. The skydiver's speed at time t (seconds AFTER he pulled the ripcord) is given by
    F= m dv/dt= -mg+ kv or dv/dt= -9.8+ (110/70)v. His height is given by the integral of v. The initial conditions are v(0)= -9.8T1 and h(0)= 700- 4.9T12.

    Solve that problem for v(t) and h(t). Solve h(t)= 0 to determine the time he hits the ground (as a function of T1) then plug that value of T1 into v(t) to determine the speed with which he hits the ground. v(t) will also depend upon T1. Finally determine T1 so that v(t)= -10.8 (which is correct).

    (Terminal velocity, using the numbers given here, is 6.86 m/s but I don't see that it has anything to do with this problem.)
     
  5. May 9, 2004 #4
    thanks HallsofIvy, much appreciative =) =), you have no idea how long ive been sitting here tryna figure that one out ... i'll try this out and see if can put it all together

    thnx again =)
     
  6. May 9, 2004 #5

    Gokul43201

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    If you pull the ripcord when your velocity is less than 6.86 m/s , then you'll never get faster than that, so you'll always have a gentle landing. So you know that the limiting T1 is definitely greater than 0.7 s. With very little calculation you have an estimate of a lower bound that may be useful for error checking.

    But you still have to solve dv/dt = 1.573v - 9.8 and find integral v(t)dt to get T1(max).
     
  7. May 10, 2004 #6
    ok im a bit confused.... just bear with me, physics i have yet to improve on...


    ok so i integrate dv/dt

    i get

    v(t) = (1.573v-9.8)t - 9.8T1 <<< CORRECT?

    or should it be

    v(t) = (1.573v-9.8)T1 - 9.8T1


    one other thing... the v value attatched to 1.573 i'm having difficulties trying to figure it out... i dont think it's -10.8 or -117.11 (free fall velocity if he actually didnt open his parachute)...

    thanks in advance
     
  8. May 12, 2004 #7

    hys

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    hey dude have you figured it out yet .. email me ... stuff@auscentral.com if you ever get it ... you should of also posted the hint that Mark put on the net...
     
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