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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?
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?

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[\Delta y=h_o+v_ot\frac12 gt^2\]\[v_1=v_0gt\] 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\[v=v_1gt\]\[\frac{vv_1}{g}=t\]this will get you the time, and also allow you to calculate the height when the acceleration goes zero

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the remaining distance covered at a constant speed is\[v=\frac{\Delta y}{\Delta t}\]\[\Delta t=\frac{\Delta y}{v}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2sum up all the time intervals to get the total time elapsed :)

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0wait, so first i'll get the delta when t=10?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2you need to keep track of all the heights and speeds along the way to keep track of the time

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2notice 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
 3 years ago
Best ResponseYou've already chosen the best response.2aside from that, we are told that at 10 seconds they open their chute and alter the equation

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2hmmm, my second setup tho is not accounting for the change in accleration is it

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2so acceleration in the second one is not gravity; but rather the 6 m/s^2

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0for the equation of delta t, what value of v will i use?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the velocity is stated as 6m/s for the final stage

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2you need to know the delta y at the end of the 2nd stage to calculate the proper time of the last stage

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0in 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
 3 years ago
Best ResponseYou've already chosen the best response.2for 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,\[v=29.8(10)\] 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
 3 years ago
Best ResponseYou've already chosen the best response.2starts at 945 meters, with a velocity of 96 \[\Delta y=\frac126t^296t+945\] we know at a certain "time" the velocity is equal to 6 m/s

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2should we have a negative in from there? or does that sign change due to the acceleration being inthe opposite direction?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2i think this might be a better construction \[\Delta y=\frac126t^296t+945\]\[v=6t96\]\[t=\frac{v+96}{6}~:~v=6\]that looks better to me

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0ah, 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
 3 years ago
Best ResponseYou've already chosen the best response.2the 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

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0ah, ok got it. thanks a lot!!!!!!!!!!!!!!!!!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2im 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

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1your directions aren't consistent @amistre64

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2lol, my thoughts arent consistent either :)

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1for deacceleration stage a is + ; Vi is  ; Vf is  ;

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2yeah, i still had reservations about that.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2so v=6 instead of 6 in that case

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0it's deceleration, and it's opposite to the initial direction of motion, then why is a +?

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1because motion is down? and you already called 'down' negative in earlier equations?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2so 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

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0in the equation that u've mentioned earlier v=6t96, 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
 3 years ago
Best ResponseYou've already chosen the best response.2you forgot to include the initial height of 945 meters that it starts at

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2ugh .... misread the post lol

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.26 = 6t 96 6 + 96 = 6t 90 = 6t 90/6 = t = 15

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0so i will add the time intervals 10, 15 and ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.210 + 15 + .... v = d/t t = d/v ; t = 180/6

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0w8, isn't the v negative? sorry, jst confused....

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.2if you are traveling at 6 m/s east, you cover the same amount of turf as if you had traveled 6m/s to the southwest

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2we 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

amrit_srovar
 3 years ago
Best ResponseYou've already chosen the best response.0ah, ok. got it. thanks @amistre64 and @Algebraic!
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