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

In summary: RIGHT answerIn summary, the two skydivers, who start from the same height and want to land on the ground at the same time, are faced with a problem of determining when the second skydiver should open her parachute. Using the equation yf=y0+(v0*t)-(.5gt^2) and assuming a constant acceleration of 32ft/sec before the parachute opens and a constant velocity of 12 ft/sec after the parachute opens, a quadratic equation must be solved to find the optimal time for the second skydiver to open her parachute. By setting the total time to land equal to 427 seconds, the
  • #1
NamaeKana
16
0

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 ?
 
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  • #2
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
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
NamaeKana said:
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
Next, I need to find T for Sky2 first, then I can calculate the free fall & chute-open times.
 
  • #6
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
NamaeKana said:
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
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
I meant

= +156 + 12T - 16T^2
 

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

1. How does the "Skydiver #1 #2 fall problem" relate to linear and quadratic equations?

The "Skydiver #1 #2 fall problem" is a classic physics problem that involves two skydivers of different weights falling from the same height. This problem can be modeled using both linear and quadratic equations depending on the variables and assumptions made.

2. What is the difference between linear and quadratic equations in the context of the "Skydiver #1 #2 fall problem"?

A linear equation in this context would involve only one variable (such as time) and a constant rate of change. On the other hand, a quadratic equation would involve multiple variables (such as time and acceleration) and a changing rate of change.

3. How do you determine which type of equation to use for the "Skydiver #1 #2 fall problem"?

The type of equation used for the "Skydiver #1 #2 fall problem" depends on the specific information given in the problem. If the problem involves a constant rate of change, a linear equation would be appropriate. If the problem involves a changing rate of change, a quadratic equation would be necessary.

4. What are some real-life applications of the "Skydiver #1 #2 fall problem" and its corresponding equations?

The "Skydiver #1 #2 fall problem" and its equations have various applications in physics and engineering, such as predicting the trajectory of projectiles and analyzing the motion of objects under the influence of gravity.

5. Are there any limitations or assumptions made when using linear or quadratic equations for the "Skydiver #1 #2 fall problem"?

As with any mathematical model, there are limitations and assumptions made when using linear or quadratic equations for the "Skydiver #1 #2 fall problem". These may include assuming a constant air resistance, neglecting the effects of wind and air currents, and assuming a perfectly symmetrical fall for both skydivers.

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