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amrit_srovar Group Title

a man jumps from a balloon which is rising vertically at a constant rate of 2 mps at an elevation of 1415 m above the ground. The man waits 10 seconds before pulling the parachute cord. the parachute decelerates him at a uniform rate of 6 mps^2 until a speed of 6mps is attained. the man then descends to the ground at this constant speed. how long will it take him to reach the ground from the time he jumps from the balloon?

  • 2 years ago
  • 2 years ago

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  1. amistre64 Group Title
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    \[\Delta y=h_o+v_ot-\frac12 gt^2\]\[v_1=v_0-gt\] at t=10, you will know the starting height and velocity of the next step\[\Delta y=h_1+v_1t-\frac12 gt^2\]determine the time by using the velocity information\[v=v_1-gt\]\[\frac{v-v_1}{-g}=t\]this will get you the time, and also allow you to calculate the height when the acceleration goes zero

    • 2 years ago
  2. amistre64 Group Title
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    the remaining distance covered at a constant speed is\[v=\frac{\Delta y}{\Delta t}\]\[\Delta t=\frac{\Delta y}{v}\]

    • 2 years ago
  3. amistre64 Group Title
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    sum up all the time intervals to get the total time elapsed :)

    • 2 years ago
  4. amrit_srovar Group Title
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    wait, so first i'll get the delta when t=10?

    • 2 years ago
  5. amrit_srovar Group Title
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    delta y*

    • 2 years ago
  6. amistre64 Group Title
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    you need to keep track of all the heights and speeds along the way to keep track of the time

    • 2 years ago
  7. amistre64 Group Title
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    notice that the first velocity you have is +2 m/s from the force of the rising balloon; this turns the inital state into a little more complex than a free falling object

    • 2 years ago
  8. amistre64 Group Title
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    aside from that, we are told that at 10 seconds they open their chute and alter the equation

    • 2 years ago
  9. amistre64 Group Title
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    hmmm, my second setup tho is not accounting for the change in accleration is it

    • 2 years ago
  10. amistre64 Group Title
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    so acceleration in the second one is not gravity; but rather the 6 m/s^2

    • 2 years ago
  11. amrit_srovar Group Title
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    for the equation of delta t, what value of v will i use?

    • 2 years ago
  12. amistre64 Group Title
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    the velocity is stated as 6m/s for the final stage

    • 2 years ago
  13. amistre64 Group Title
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    you need to know the delta y at the end of the 2nd stage to calculate the proper time of the last stage

    • 2 years ago
  14. amrit_srovar Group Title
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    in solving for the delta y, i will just use the given equation right? i will use it in the given where t=10,h=1415;v=2; am i right?

    • 2 years ago
  15. amistre64 Group Title
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    for the 1st stage yes; and g=9.8\[\Delta y=-4.9(10^2)+2(10)+1415\]this gives us the starting height of the 2nd stage, but we also need the starting velocity of the second stage,\[v=2-9.8(10)\] this tells us what we would need to do to rewrite the equation, with the added information that the acceleration is now 6 instead of 9.8

    • 2 years ago
  16. amistre64 Group Title
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    starts at 945 meters, with a velocity of -96 \[\Delta y=-\frac126t^2-96t+945\] we know at a certain "time" the velocity is equal to 6 m/s

    • 2 years ago
  17. amistre64 Group Title
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    should we have a negative in from there? or does that sign change due to the acceleration being inthe opposite direction?

    • 2 years ago
  18. amistre64 Group Title
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    i think this might be a better construction \[\Delta y=\frac126t^2-96t+945\]\[v=6t-96\]\[t=\frac{v+96}{6}~:~v=6\]that looks better to me

    • 2 years ago
  19. amrit_srovar Group Title
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    ah, so i'll use the v=6 to compute for t, which i'll use in the two upper equations in ur recent post. after this, i'll use the v and delta y to get delta t. so is the delta t the final answer?

    • 2 years ago
  20. amistre64 Group Title
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    the sum of all the time intervals is the answer. 10 seconds in the first stage 17 seconds in the second stage and what, 180/6 for the final stage

    • 2 years ago
  21. amrit_srovar Group Title
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    ah, ok got it. thanks a lot!!!!!!!!!!!!!!!!!

    • 2 years ago
  22. amistre64 Group Title
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    im still a bit leary about my thoughts for the second stage. im pretty sure i read it right and that acceleration is 6m/s^2 in the upward direction

    • 2 years ago
  23. Algebraic! Group Title
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    your directions aren't consistent @amistre64

    • 2 years ago
  24. amistre64 Group Title
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    lol, my thoughts arent consistent either :)

    • 2 years ago
  25. Algebraic! Group Title
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    for deacceleration stage a is + ; Vi is - ; Vf is - ;

    • 2 years ago
  26. amistre64 Group Title
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    yeah, i still had reservations about that.

    • 2 years ago
  27. amistre64 Group Title
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    so v=-6 instead of 6 in that case

    • 2 years ago
  28. Algebraic! Group Title
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    yep

    • 2 years ago
  29. amrit_srovar Group Title
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    it's deceleration, and it's opposite to the initial direction of motion, then why is a +?

    • 2 years ago
  30. Algebraic! Group Title
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    because motion is down? and you already called 'down' negative in earlier equations?

    • 2 years ago
  31. amrit_srovar Group Title
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    ah, yeah. thnx

    • 2 years ago
  32. Algebraic! Group Title
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    np:)

    • 2 years ago
  33. amistre64 Group Title
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    so out time is 15 in the second stage, which also alters the final distance to travel 3(15^2)-96(15)+945 = 180 , ironically 16 is the vertex of the parabola, so 17 = 16 .... final distance is still 180

    • 2 years ago
  34. amistre64 Group Title
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    y(17)=y(15) that is

    • 2 years ago
  35. amrit_srovar Group Title
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    in the equation that u've mentioned earlier v=6t-96, since the new t is 15, if we're going to substitute 15 in this equation, the answer is negative, thus the v is negative. when we r going to substitute the v in delta t=delta y/v: delta t=180/-6, the answer is negative....?

    • 2 years ago
  36. amistre64 Group Title
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    you forgot to include the initial height of 945 meters that it starts at

    • 2 years ago
  37. amistre64 Group Title
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    ugh .... misread the post lol

    • 2 years ago
  38. amistre64 Group Title
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    -6 = 6t -96 -6 + 96 = 6t 90 = 6t 90/6 = t = 15

    • 2 years ago
  39. amrit_srovar Group Title
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    so i will add the time intervals 10, 15 and ?

    • 2 years ago
  40. amistre64 Group Title
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    the new height at t=15 from the second stage is 3(15^2)-96(15)+945 = 180 for the final height of that stage, which is the initial height for the last stage

    • 2 years ago
  41. amistre64 Group Title
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    10 + 15 + .... v = d/t t = d/v ; t = 180/6

    • 2 years ago
  42. amrit_srovar Group Title
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    w8, isn't the v negative? sorry, jst confused....

    • 2 years ago
  43. amistre64 Group Title
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    the v and the distance are both negative in the final stage if viewed from a certain vantage point. but all we are really concerned about there is that the distance traveled at the speed allotted. speed is independant of direction traveled

    • 2 years ago
  44. amistre64 Group Title
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    if you are traveling at 6 m/s east, you cover the same amount of turf as if you had traveled 6m/s to the south-west

    • 2 years ago
  45. amistre64 Group Title
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    we have 180 meters left to travel at a speed of 6 m/s if we want that in vectors; we have a -180 meter displacement to cover by a -6 m/s velocity

    • 2 years ago
  46. amrit_srovar Group Title
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    ah, ok. got it. thanks @amistre64 and @Algebraic!

    • 2 years ago
  47. amistre64 Group Title
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    youre welcome :)

    • 2 years ago
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