- #1
sid9221
- 111
- 0
http://dl.dropbox.com/u/33103477/parachutist.png
(This is from a mathematics not physics exam)
Well what I've done, is to compose the equation describing the velocity of the parachute once it has opened. I did this by setting up the differential equation as:
[tex] \frac{dv}{dt} = g-kv [/tex]
I solved that to get:
[tex] v = \frac{g}{k} - \frac{Ce^{-kt}}{k} [/tex]
Now the parachutist was traveling at v_0 when it deployed so I can say V(0)=v_0
hence solving I got:
[tex] v(t)= \frac{g}{k} - \frac{g-v_0k}{k}e^{-kt} [/tex]
Now the point is If I know how far from the ground the parachute opened I can figure out the distance traveled since the parachute was opened ?
I'm confused so any direction on how I should get to x(t) from v(t) would be useful.
(This is from a mathematics not physics exam)
Well what I've done, is to compose the equation describing the velocity of the parachute once it has opened. I did this by setting up the differential equation as:
[tex] \frac{dv}{dt} = g-kv [/tex]
I solved that to get:
[tex] v = \frac{g}{k} - \frac{Ce^{-kt}}{k} [/tex]
Now the parachutist was traveling at v_0 when it deployed so I can say V(0)=v_0
hence solving I got:
[tex] v(t)= \frac{g}{k} - \frac{g-v_0k}{k}e^{-kt} [/tex]
Now the point is If I know how far from the ground the parachute opened I can figure out the distance traveled since the parachute was opened ?
I'm confused so any direction on how I should get to x(t) from v(t) would be useful.
Last edited by a moderator: