Question

Any ideas on what to do with this?

Answer 1

good lord no
is this a diffeq problem?

Answer 2

yep :)

Answer 3

I await someone who knows how to answer this.

Answer 4

glad it is you and not me. sorry

Answer 5

i dont think weight will affect acceleration due to gravity

Answer 6

nah, but it will have some affect on terminal velocity

Answer 7

once the chute opens; do you still factor in the weight of the parachute?

Answer 8

and once it opens; do you go back to v(0) = 0 ?

Answer 9

or .... is the new v(0) the v(15) from the first go around ......

Answer 10

\[m\frac{dv}{dt}=mg-kv\]

\[\frac{dv}{dt}+\frac{k}{m}v=g\]

\[exp(\frac{k}{m}t)\ v=\frac{mg}{k}\int \frac{k}{m}exp(\frac{k}{m}t)\ dt\]

\[exp(\frac{k}{m}t)\ v=\frac{mg}{k}exp(\frac{k}{m}t)+C_1\]

\[v(t)=\frac{mg}{k}+C_1exp(-\frac{k}{m}t)\]

\[m=72.57kg;\ k=.5;\ g=9.8m/s^2\]

\[v(0)=0=\frac{72.57(9.8)}{.5}+C_1\]

\[v(t)=1422.372 (1-exp(-.0069\ t))\]

\[v(15)=1422.372 (1-exp(-.0069\ (15)))=139.853\]

Answer 11

now, i wanna say we use this v(15) value for the new initial value condition on the open parachute;

Answer 12

i know if I integrate velocity; the area under the curve is a measure of displacement so I can find how far they have dropped

Answer 13

\[1422.372\int _{0}^{15} (1-exp(-.0069\ t))\ dt\]

\[1422.372(t+\frac{exp(-.0069\ t)}{.0069}) \]

\[1422.372(15+\frac{exp(-.0069(15))}{.0069})-1422.372(\frac{1}{.0069})=1066.99\ meters \]
thanks to the wolf lol

Answer 14

\[v(t)=\frac{mg}{k}+C_1exp(-\frac{k}{m}t)\]
\[m=72.57kg;\ k=10;\ v(0)=139.853\]
\[\frac{72.57(9.8)}{10}=71.1186;\ -\frac{10}{72.57}=-0.1378\]
\[v(0)=139.853=71.1186+C_1;\ C_1=68.7344\]
\[v(t)=71.1186+68.7344exp(-0.1378\ t)\]
would be our new equation to play with

Answer 15

5 more seconds = 20 seconds total sooo
\[v(5)=71.1186+68.7344exp(-0.1378(5))=105.629\]
so it least shes slowing down :)

Answer 16

and the integration of that from 0 to 5 gets me .... 603.956 meters

so she should have traveled a distance of:

1066.99
603.956
---------
1670.946 meters or so .....

5483 feet if we are lucky :)

Answer 17

just need to obtain a terminal velocity to compare with; and also how long it takes to land

Answer 18

is terminal velocity with the chute open? or closed?

they really need to provide you with all this pertinent information ....

Answer 19

terminal velocity is at 0 acceleration; so derivative again to determine acceleration:

\[v(t)=\frac{mg}{k}+C_1exp(-\frac{k}{m}t)\]

\[a(t)=-\frac{k}{m}\ 68.7344\ exp(-\frac{k}{m}t)\]
with chute open
\[a(t)=-\frac{k}{m}\ 1422.372\ exp(-\frac{k}{m}t)\]
with chute NOT open :)

hmmm, none of these go to zero in order to determine a zero acceleration tho

Answer 20

i might as well use the v' from the onset since thats how this stuff was defined; and i might be able to get a zero from that:

\[v'=mg-kv\]
\[v'=72.57(9.8)- 711.186-687.344\ exp(−0.1378 t)=0\]
still no luck

Answer 21

\[v(t)=1422.372\ (1-exp(-\frac{.5t}{72.57}))\]
\[v(15)=139.6589\]

Answer 22

\[v(t)=71.1186+68.5403\ exp(-\frac{10}{72.57}\ t)\]

Answer 23

\[v(5) = 105.5315 \]