Ya, thats really good. If it was slightly conical it would definitely stop the egg from slipping down on impact. The paper in a conical shape wouldnt b quite as strong as cylindrical though
U dont have or need a time variable so thats the wrong equation of motion. This 1 shud work better:
v^2 = v_0^2 + 2ax
The other 1s r listed here if u need em:
http://en.wikipedia.org/wiki/Equations_of_motion
Well, u got the right answer for v_terminal anyway in ur 1st post. (v^2 = 13.06)
U still need more info about before the parachute was opened to get the drag just after.
The drag force u calculated (783. 6) is the drag force when the skydiver achieved v_terminal, which is a little while after...
How do u no ur wrong? Looks ok 2 me unless im makin exactly the same mistake. Mayb ur lookin under the wrong chapter at the back of the book 4 ur answers!?
Well u just hav 2 use the same equations of motion
The usual convention for these 1s is up is positive and down is negative so the initial speed u is a positive num since in the up direction. The displacement s will be -100 and g is -9.81. Putting in the numbers:
s = u*t - 0.5*g*t^2
-100 =...