Question

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?

Answer 1

\[\Delta y=h_o+v_ot-\frac12 gt^2\]\[v_1=v_0-gt\]
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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

delta y*

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

your directions aren't consistent @amistre64

Answer 24

lol, my thoughts arent consistent either :)

Answer 25

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

Answer 26

yeah, i still had reservations about that.

Answer 27

so v=-6 instead of 6 in that case

Answer 28

yep

Answer 29

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

Answer 30

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

Answer 31

ah, yeah. thnx

Answer 32

np:)

Answer 33

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

Answer 34

y(17)=y(15) that is

Answer 35

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....?

Answer 36

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

Answer 37

ugh .... misread the post lol

Answer 38

-6 = 6t -96

-6 + 96 = 6t

90 = 6t

90/6 = t = 15

Answer 39

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

Answer 40

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

Answer 41

10 + 15 + ....

v = d/t

t = d/v ; t = 180/6

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

youre welcome :)