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Skydiver #1 #2 fall problem, Linear or Quadratic ?

  • Thread starter NamaeKana
  • Start date
  • #1
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Homework Statement



Skydiver #1 from the University Skydiving Club steps out of a plane when it is 1 miles above the ground and 10 seconds later skydiver #2 steps out of a plane (at the same height).

They both want to land on the ground at the same time.

To make an estimate assume that a skydiver falls with a constant acceleration of 32ft/sec before the parachute opens and once the parachute has opened the skydiver falls with a constant velocity of 12 ft/sec.

If skydiver #1 opens his parachute 2 seconds after stepping out of the plane, how long should skydiver #2 wait before opening her parachute ?

(You may need to solve a quadratic equation)


Homework Equations



yf=y0+(v0*t)-(.5gt^2)


The Attempt at a Solution



i just need a few hints. i think i can solve it.

is this a linear problem or a quadratic problem ?
 

Answers and Replies

  • #2
1,930
229
1 Compute after how many seconds skydiver #1 reaches the ground.
- compute how far skydiver #1 falls in the first 2 seconds. this is a quadratic problem
- compute the time needed for the rest of the fall. This is a linear problem

2 Now do the same for skydiver #2 only use T seconds instead of 2 seconds.

since skydiver #2 starts 10 seconds later, the time found in step 2 should be ......
 
  • #3
16
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from 0-2 sec (.5 x 32 x 2^2)= 64 ft
5280 ft - 64 ft = 5216 ft
5216 ft / 12 ft/s = 435 sec
so sky1 takes 435 + 2 = 437 secs to hit ground

sky2 jumps 10 secs later 437 - 10 = 427 secs

5280 = 0 + ( 12 x (427 - T)) + (.5 x 32 T^2)

the last quadratic didn't work. hmm...
 
  • #4
1,930
229
from 0-2 sec (.5 x 32 x 2^2)= 64 ft
5280 ft - 64 ft = 5216 ft
5216 ft / 12 ft/s = 435 sec
so sky1 takes 435 + 2 = 437 secs to hit ground

sky2 jumps 10 secs later 437 - 10 = 427 secs

5280 = 0 + ( 12 x (427 - T)) + (.5 x 32 T^2)

the last quadratic didn't work. hmm...
You must do the same with sky2 as you did with sky1.
First calculate the distance fallen in the first T seconds.
then calculate the time needed for the rest of the distance to the ground.
Then the sum of these times for these 2 parts of the trajectory must equal 427s
 
  • #5
16
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Next, I need to find T for Sky2 first, then I can calculate the free fall & chute-open times.
 
  • #6
16
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i manually figured out that for Sky#2 to fall 5280 ft, it takes T = 3.5 secs

5280 = 12(427 - T) + (16 * T^2)

next i tried with variable T

5280 = 12(427 - T) - 16T^2)
5280 = 5124 - 12T -16T^2
= -156 -12T -16T^2

the root is -.375, and it's wrong

but if i manually change it to

= +156 -12T -16T^2

the roots are -3.51, +2.77, |-3.51| is the RIGHT answer

i'm still trying to get it right w/o fudging stuff.....
 
  • #7
1,930
229
Next, I need to find T for Sky2 first, then I can calculate the free fall & chute-open times.
A technique that is uses in many physics or math problems is:

If you don't know what something is, call it x and see what happens
In this case you calculate these times as a function of T and then you find T using the fact that the total time to land must be equal to 427s
 
  • #8
16
0
O.K. I got it now, thanks for your help ;-)

-------------------------------------------------------
Yf= Yo + VoyT - .5gT^2
0 = 5280 - 12(427 - T) - 16T^2
= 5280 - 5124 + 12T - 16T^2
= -156 + 12T - 16T^2

the roots are -2.77, +3.51, and 3.51 is the RIGHT answer
-------------------------------------------------------
 
  • #9
16
0
I meant

= +156 + 12T - 16T^2
 

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