anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • 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
amistre64
  • amistre64
the remaining distance covered at a constant speed is\[v=\frac{\Delta y}{\Delta t}\]\[\Delta t=\frac{\Delta y}{v}\]
amistre64
  • amistre64
sum up all the time intervals to get the total time elapsed :)

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anonymous
  • anonymous
wait, so first i'll get the delta when t=10?
anonymous
  • anonymous
delta y*
amistre64
  • amistre64
you need to keep track of all the heights and speeds along the way to keep track of the time
amistre64
  • 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
amistre64
  • amistre64
aside from that, we are told that at 10 seconds they open their chute and alter the equation
amistre64
  • amistre64
hmmm, my second setup tho is not accounting for the change in accleration is it
amistre64
  • amistre64
so acceleration in the second one is not gravity; but rather the 6 m/s^2
anonymous
  • anonymous
for the equation of delta t, what value of v will i use?
amistre64
  • amistre64
the velocity is stated as 6m/s for the final stage
amistre64
  • amistre64
you need to know the delta y at the end of the 2nd stage to calculate the proper time of the last stage
anonymous
  • anonymous
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?
amistre64
  • 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
amistre64
  • 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
amistre64
  • amistre64
should we have a negative in from there? or does that sign change due to the acceleration being inthe opposite direction?
amistre64
  • 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
anonymous
  • anonymous
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?
amistre64
  • 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
anonymous
  • anonymous
ah, ok got it. thanks a lot!!!!!!!!!!!!!!!!!
amistre64
  • 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
anonymous
  • anonymous
your directions aren't consistent @amistre64
amistre64
  • amistre64
lol, my thoughts arent consistent either :)
anonymous
  • anonymous
for deacceleration stage a is + ; Vi is - ; Vf is - ;
amistre64
  • amistre64
yeah, i still had reservations about that.
amistre64
  • amistre64
so v=-6 instead of 6 in that case
anonymous
  • anonymous
yep
anonymous
  • anonymous
it's deceleration, and it's opposite to the initial direction of motion, then why is a +?
anonymous
  • anonymous
because motion is down? and you already called 'down' negative in earlier equations?
anonymous
  • anonymous
ah, yeah. thnx
anonymous
  • anonymous
np:)
amistre64
  • 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
amistre64
  • amistre64
y(17)=y(15) that is
anonymous
  • anonymous
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....?
amistre64
  • amistre64
you forgot to include the initial height of 945 meters that it starts at
amistre64
  • amistre64
ugh .... misread the post lol
amistre64
  • amistre64
-6 = 6t -96 -6 + 96 = 6t 90 = 6t 90/6 = t = 15
anonymous
  • anonymous
so i will add the time intervals 10, 15 and ?
amistre64
  • 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
amistre64
  • amistre64
10 + 15 + .... v = d/t t = d/v ; t = 180/6
anonymous
  • anonymous
w8, isn't the v negative? sorry, jst confused....
amistre64
  • 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
amistre64
  • 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
amistre64
  • 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
anonymous
  • anonymous
ah, ok. got it. thanks @amistre64 and @Algebraic!
amistre64
  • amistre64
youre welcome :)

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