## amrit_srovar 3 years ago 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?

1. amistre64

$\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. amistre64

the remaining distance covered at a constant speed is$v=\frac{\Delta y}{\Delta t}$$\Delta t=\frac{\Delta y}{v}$

3. amistre64

sum up all the time intervals to get the total time elapsed :)

4. amrit_srovar

wait, so first i'll get the delta when t=10?

5. amrit_srovar

delta y*

6. amistre64

you need to keep track of all the heights and speeds along the way to keep track of the time

7. amistre64

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

8. amistre64

aside from that, we are told that at 10 seconds they open their chute and alter the equation

9. amistre64

hmmm, my second setup tho is not accounting for the change in accleration is it

10. amistre64

so acceleration in the second one is not gravity; but rather the 6 m/s^2

11. amrit_srovar

for the equation of delta t, what value of v will i use?

12. amistre64

the velocity is stated as 6m/s for the final stage

13. amistre64

you need to know the delta y at the end of the 2nd stage to calculate the proper time of the last stage

14. amrit_srovar

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?

15. amistre64

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

16. amistre64

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

17. amistre64

should we have a negative in from there? or does that sign change due to the acceleration being inthe opposite direction?

18. amistre64

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

19. amrit_srovar

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?

20. amistre64

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

21. amrit_srovar

ah, ok got it. thanks a lot!!!!!!!!!!!!!!!!!

22. amistre64

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

23. Algebraic!

your directions aren't consistent @amistre64

24. amistre64

lol, my thoughts arent consistent either :)

25. Algebraic!

for deacceleration stage a is + ; Vi is - ; Vf is - ;

26. amistre64

yeah, i still had reservations about that.

27. amistre64

so v=-6 instead of 6 in that case

28. Algebraic!

yep

29. amrit_srovar

it's deceleration, and it's opposite to the initial direction of motion, then why is a +?

30. Algebraic!

because motion is down? and you already called 'down' negative in earlier equations?

31. amrit_srovar

ah, yeah. thnx

32. Algebraic!

np:)

33. amistre64

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

34. amistre64

y(17)=y(15) that is

35. amrit_srovar

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....?

36. amistre64

you forgot to include the initial height of 945 meters that it starts at

37. amistre64

ugh .... misread the post lol

38. amistre64

-6 = 6t -96 -6 + 96 = 6t 90 = 6t 90/6 = t = 15

39. amrit_srovar

so i will add the time intervals 10, 15 and ?

40. amistre64

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

41. amistre64

10 + 15 + .... v = d/t t = d/v ; t = 180/6

42. amrit_srovar

w8, isn't the v negative? sorry, jst confused....

43. amistre64

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

44. amistre64

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

45. amistre64

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

46. amrit_srovar

ah, ok. got it. thanks @amistre64 and @Algebraic!

47. amistre64

youre welcome :)